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 IN MEMOR1AM 
 FLOR1AN CAJOR1 
 
IMPROVED EDITION WITH QUESTIONS. 
 
 A 
 
 SHORT SYSTEM OF 
 
 PRACTICAL, ARITHMETIC, 
 
 COMPILED 
 
 FROM THE BEST AUTHORITIES,; 
 
 f TO WHICH IS ANNEXED 
 
 A SHORT PLAN OF BOOK-KEEPING. 
 
 * THE WHOLE DESIGNED 
 
 **> 
 
 .' i FOR TEE USE OF SCHOOLS. ~& 
 
 BY WILLIAM KINNE, A. M. ! ' 
 
 asirftfon, 
 
 WITH QUESTIONS 
 
 ON EVERY PART OF ARITHMETIC, AND A COMPENDIOUS 
 SYSTEM OF TAX MAKING. 
 
 REVISED, CORRECTED, AND GREATLY ENLARGED, 
 
 BY DANIEL ROBINSON. 
 
 HA LL O WELL : 
 
 PRINTED AWD PUBLISHED BY GLAZIER, MASTERS & Co. 
 
 Sold by them at the Hailowell Bookstore, No. I, Kennebec-Row , and by all 
 
 the Booksellers in the State. 
 
 1831. 
 
DISTRICT OF MAINE, si. 
 
 BE IT REMEMBERED, That on the seventh day of December, in the 
 year of our Lord one thousand eight hundred and twenty-two, and the forty - 
 seventh year of the Independence of the- United States of America, GOOD- 
 ALE, GLAZIER & COMPANY of the District of Maine, have deposited in this 
 office, the title of a Book, the right whereof they claim as proprietors in the 
 words following, viz. : " Improved edition with questions. A short system 
 ' of Practical Arithmetic, compiled from the best authorities ; to which is 
 1 annexed a short plan of Book-keeping. The whole designed for the use of 
 ' Schools. By William Kinne, A, M. Fourth edition, with qaestions on every 
 * part of Arithmetic, and a compendious system of Tax making. Revised, 
 ' corrected and greatly enlarged, by Daniel Robinson. Hallowell, printed 
 ' and published 'by Goodale, Glazier & Co." In conformity to the Act of 
 the Congress of the United-States, entitled, " An Act for the encouragement 
 of learning, by securing the copies of maps, charts, and books, to the authors 
 aad proprietors of such copies, during the times therein mentioned j and also 
 to an Act, entitled " An Act supplementary to an Act, entitled an Act for the 
 encouragement of learning, by securing the copies of maps, charts and books 
 to the authors and proprietors of such copies, during the times therein men- 
 tioned, and extending the benefits thereof to the arts of designing, engraving, 
 arid etching historical and other prints." 
 
 JOHN MUSSEY, JR., Clerk of the District Court of Maine. 
 
 A true copv as of record. 
 
 Attest, J. MUSSEY, JR., Clerk D. C. Maine. 
 
 Advertisement to the Seventh Edition. 
 
 J has been the primary purpose, in each improved Edition of this Work, 
 to render it more and more plain and practical, while it should embrace every 
 useful rule and question which might occur in the ordinary business transac- 
 tions of life. To effect this object, neither time nor thought has been, in any 
 wise, niggardly expended. Whatever was judged to be wanted, to charac- 
 terize it as a plain, practical, and useful system, has been amply, though 
 gradually, supplied. In the edition now presented to the public, part of the 
 questions in which avoirdupois weight is concerned, has been writtenjanew, or 
 so altered as to allow 25 pounds only to the quarter of a hundred weight ; 
 because this practice now generally obtains in business, among merchants 
 and traders in the United-States, and has moreover been established in Maine 
 by legislative enactment. Considerable new matter also has been crowd- 
 ed into the volume, and a small portion of the old withdrawn. Errour has 
 been diligently sought for and corrected; and, it is confidently believed, is 
 now nowhere to be found on its pages. Considered as an Epitome, whether 
 it be susceptible of any farther degree of improvement, may be reasonably v 
 ^ questioned. The hope is, therefore, indulged, that, though the tongue of the j) / 
 
 /7/Ycaptious caviller should blazon defects for which others might search in vain ; 
 
 C*i// vet the eye of the candid critic will see nothing in this compendium which 
 reason and truth would long hesitate to .approve. 
 Gardiner, August 1, 1828. 
 
To adapt tliis work to the easy use of Instructors, I have endeavoured to 
 simplify the definitions and rules, so as to render them as familiar and concise 
 as the nature of the subject admits. At the san-ie time, I have very consider- 
 ably enlarged the Original System, by the insertion of a far greater number 
 of practical examples, especially in the grourid-ruies, and by the introduction 
 of many new rules, in order to furnish our Schools with a methodical and 
 comprehensive Treatise of Practical Arithmetic. 
 
 Works of this kind have too often abounded with abstruse and intricate 
 questions, more puzzling than beneficial to the learner. And some authors 
 have dwelt too much on those of a trifling 1 nature, which,, when understood, 
 afford no useful knowledge. To avoid these extremes, to feed and invigorate 
 the mind, and thus form onr youth for entering, with fair promise, on the pur- 
 suits of active life, have been my principal aims ; in preparing this edition for 
 the press. 
 
 Most of the former demonstrations have been omitted, as being little suited 
 to enlighten the pupil, and as excluding, in such compends, matter much more 
 conducive to the purpose of his instruction. The book-keeping, also, has 
 been somewhat abridged, for the admissioi\of other matter j yet enough, it is 
 conceived, has been retained to give the student no very imperfect idea of this 
 branch of learning. Besides what has been substituted in place of this excluded 
 matter, no fewer than 51 pages have been added to the last edition. To the 
 whole have been prefixed brief questions on all the most important parts of 
 Arithmetic. But, instead of entering into a detail of these enlargements, I 
 beg leave to refer the reader to the table of contents, or to the pages of the 
 work itself. 
 
 During the many }'ears that I have devoted to the instruction of youth in 
 Arithmetic, 1 have used various systems, all of which havejust claims to sci- 
 entific merit. The authors, however, have, generally, appeared to be defi- 
 cient in an important point the practical teacher's experience. They have , 
 ( Jlgg^^lJiClLtQQ- spa ring of examples^more^ esneciajjy in the A^^Biles^ __The JLL&Ll 
 [ coisequeSce"!^ tfiaTlnVsch^Dlaris^lmrried ihroug ; h i me^eTundarnenta 1 rules fas- \J !*' ^-- 
 ter than his comprehension and proficiency would justify. To obviate this ob- 
 jection, has been another design in the present undertaking. 
 
 Considering that, to attain a thorough knowledge of vulgar fractions, is 
 usually too difficult a task for young students, whose progress in Arithmetic 
 has extended only to compound division, and that the difficulty frequently re- 
 sults in their utter discouragement j I have, therefore, deemed it most advise- 
 abie and advantageous to transfer these fractions (except two or three prob- 
 lems .jntrod^tor^to_dgcnals) beyond equation o* payments.^ But, asjfocimat 
 fracTibns'may be more easily acquired, are more simple, useful, ana necessa-' 
 ry, and ara sooner wanted in the practical branches of numbers, 1 have 
 thought it expedient to let them occupy that part of the work which they did 
 in former editions. The other rules I have likewise aimed so to arrange, as 
 to give precedence to those which are most simple and necessary, introducing 
 the more abstruse and difficult parts last. The teacher, however, will not 
 consider himself as being obliged to adhere strictly to this arrangement. He 
 can, notwithstanding, take the rules in such order as he may conceive to be 
 the most proper. D. R. 
 
TABLE OF CONTENTS. 
 
 Notation or Numeration, 
 r Addition, 
 Subtraction, 
 
 Multiplication, 
 Division, 
 Practical Questions, 
 
 Simple 
 
 Tables 
 
 Of 
 
 and Measures, 
 Reduction, 
 
 Application of Reduction, 
 Federal Monoy, 
 Reduction of Federal Money, 
 Practical Questions in ) 
 Federal Money, J 
 Addition, 
 
 Division, 
 Average Judgement, 
 Duodecimals, 
 . Three Problems in Vulgar 
 
 Fractions, 
 Decimal Fractions, 
 Reduction of Currencies, 
 Rule of Three, 
 Practice, 
 Tare and Tret, 
 Double Rule of Three, 
 Conjoined Proportion, 
 Barter, 
 Loss and Gain, 
 
 Simple Interest, 
 Short Practical Rules, 
 To compute Interest on ) 
 
 Notes, Bonds, &c. 
 Compound Interest, 
 Commission, 
 Insurance, 
 Discount, 
 
 Annuities at Simple Interest, 
 Equation of Payments, 
 Exchange, 
 
 PAGE\ 
 
 
 PAGE. 
 
 9 
 
 12 
 
 Vulgar Fractions, 
 Rule of Three in Vul. Frac. 
 
 147 
 160 
 
 15 
 
 Rule of Three in Dec. Frac. 
 
 161 
 
 16 
 
 Double Rule of Three in ) 
 
 _ 
 
 21 
 
 Vulgar Fractions, $ 
 
 loZ 
 
 25 
 
 Simple Interest in Decimals, 
 
 163 
 
 ? 07 
 
 Table of Ratios, 
 
 163 
 
 ) 
 
 Compound Interest in Decima 
 
 a, 167 
 
 32 
 
 Table of the amount of l, 
 
 ) 
 
 37 
 
 or $1, at 5 and 6 per cent. 
 
 / 
 
 38 
 
 per annum, compound inter- 
 
 \ 
 
 44 
 
 est, for 20 3 T ears, 
 
 ; 
 
 48 
 
 Annuities at compound interest 
 Involution, 
 
 , 168 
 169 
 
 48 
 
 Evolution, or Extraction ) 
 
 17fl 
 
 56 
 
 of Roots, 5 
 
 I / U 
 
 61 
 
 66 
 
 Application and Use of the ) 
 Square Root. \ 
 
 173 
 
 72 
 
 The Cube Root, 
 
 177 
 
 73 
 
 Application and Use of the ) 
 
 
 f. 
 
 Cube Root, 5 
 
 
 7o 
 
 Roots of Powers in general, 
 
 179 
 
 79 
 88 
 
 Progression ^ Geolnet^ical 1 ' 
 
 181 
 
 183 
 
 98 
 
 C TVTftrJJol 
 
 187 
 
 J05 
 
 Alligation , } Alternate 
 
 188 
 
 108 
 
 p . . ( Single, 
 
 190 
 
 111 
 
 Position < i) ou ble, 
 
 191 
 
 114 
 
 Permutation and Combination, 
 
 193 
 
 115 
 
 Miscellaneous Questions, 
 
 196 
 
 117 
 
 Measurement of Grindstones, 
 
 201 
 
 119 
 
 122 
 
 Mensuration < ^ sFds ^^ 
 
 202 
 211 
 
 123 
 
 Cask Gauging, 
 
 219 
 
 127 
 
 Ships' Tonnage and Length > 
 of Masts, J 
 
 221 
 
 130 
 
 Weight of Anchor a Cable \ 
 
 
 135 
 
 may sustain Burthen of f 
 
 999 
 
 137 
 
 Ships from their Propor- t 
 
 % 
 
 137 
 
 tions, 
 
 
 138 
 
 Wood and Bark Measure, 
 
 223 
 
 140 
 
 Assessing Taxes, 
 
 223 
 
 141 
 
 142 
 
 Book-keeping 1 , 
 Tables of value of gold, 
 
 226 
 236 
 
Topics for Examination in the Arithmetic. 
 
 What is Arithmetic? What are the four fundamental Rules for its opera- 
 tion ? To understand these, what is previously necessary ? \Vhat does No- 
 tation teach 1 How many characters or figures are employed in it ? By 
 what common term are the first nine called ? How named ? What does the 
 tenth figure denote ? Have not these digits a local, as well as a simple value 1 
 On what principle, does their local value depend ? . Denominate the names of 
 the places, according to their order. How is the cipher used, in connection 
 with the significant figures ? In what manner are large numbers divided? 
 Name the places * and read each line of numbers, in the Numeration Table. 
 What is the Rule for expressing numbers m figures when above nine ? Give 
 the Rule to 'read numbers. What is Addition? Simple? Let me examine 
 you in the Table. Recite the Rule and modes of Proof. What is Subtrac- 
 tion? When Simple? Name its numbers. Let me examine you in the Ta- 
 ble. Give the Rule and way of Proof. What is Multiplication ? Name its 
 numbers What common term is applied to \\\G first two of them 1 When is 
 it Simple? How is the Table used ? Let me examine you in it. When is it 
 Cast first? Repeat the Rule. When Case second? Give the Rule, and 
 modes of Proof. When does the first Case of Contractions apply 1 Give the 
 Rule. When the second Case ? Tel) the Rule. What is shown by Divi- 
 sion 1 Name its numbers. When is it Simple ? Let me examine you in the 
 Table. Repeat the Rule, Notes, and modes of Proof. Repeat the directions 
 in Case first of Contractions. What Case second and Rule 1 What says 
 Case third ? What its Note and Rule ? Repeal the Money Table. That of 
 Troy Weight. Apothecaries' Weight. Avoirdupois Weight. Cloth Meas- 
 ure. Long; Measure. Square Measure. Cubic Measure, Dnj Measure. 
 Wine Measure. Ale Measure. Time. Planetary Motion. .What is taught 
 by Reduction 1 Repeat the Rules and Proof. Tell me the mode of for- 
 mation,' and Tabled Federal Money. Give the Ride for its Addition For 
 its Subtraction. For its Multiplication. Recite ihe J\'ote, and A Short Rn/e. 
 Give the Rule for its Division. Tell the Short Rule. Give the Rules for the 
 Reduction of Federal Money. What is the DiiecLion in Case first for chang- 
 ing New-England currency to Federal Money ? In case second 1 In Case 
 third? What is the Ride in Case first for "changing Federal Money to 
 New-England currency ? In Case second ? Whk is taught by Compound 
 Addition? Give the Rule. What is Compound) fubf.raclf.on? Toll the 
 Rule. What does Compound Multiplication teach ? Give the Rule. What 
 is the direction in Case second 1 What in Cae third 1 What does Com- 
 pound Division teach ? Rehearse the Rule. What does Case second direct ? 
 What Case tfu'rrf ? Tell the JVbte before examples in Average Judgment, 
 What is observed of Duodecimals 1 Give the Rule for multiplying them. 
 What are Fractions ? How is a Vulgar Fraction represented? Name its 
 parts. What is shown by the Denominator ? What by the Numerator? 
 When is a fraction in its lowest terms? Give the rule' for Problem first. 
 What is the intent of Problem second ? Tell its rule. What of Problem 
 third 1 What \isrule? What is a Decimal Fraction? How expressed ? 
 What determines its relative value ? How are such fractions affected by ci~ 
 
TI TOPICS FOR EXAMINATION. 
 
 Reduction ? In second Case ? IB third 1 In fourth 1 In fifth ? What are 
 the rides in Cascjirst of Reduction of Currencies ? In case'second ? third ? 
 fourth? fifth? What are the rules for changing Federal Money into the 
 Currencies of the several Slates ? What for changing it to Canada and Nova 
 Scotia money ? What to that of Great Britain ? What does the Rule of 
 Three teach ? Why so na?/^ ? Why called Golden Rule ? Give the rules, 
 and notes for its operation. What is Practice ? What the rule in Case first 1 
 What says Case second, and what its rule 1 What are Tare and 7Ve ?" De- 
 fine all its terms. When is Casejirst used, and what its rule 1 When Case 
 second, and what the rule 1 Case third, and what the rule ? Fourth, and 
 what the rule 1 What does the Double Rule of Three teach ? How many 
 terms in its questions ? How distinguished ? Give the rule and wtftes. What 
 is Conjoined Proportion ? When is Case first used, and what its rules ? 
 When Case second, and what its rule 1 What is Barter ? Its rule? What 
 Loss and Gain? In what instruct Merchants and traders? How its #z/es- 
 /zorts solved ? What its general law ? What is Fellowship 1 Its wse? What 
 Single Fellowship ? Its 72z//e ? How /?rweJ ? What Double Fellowship ? 
 Its Rule 1 What is Interest ? What the fcg-^ interest ? Define its term. 
 How many kinds ? What Simple? ]is Rule and Note ? Tell the 7\iW<? of 
 Aliquot Farts '? What the Rule for Months, at 6 per cent. ? For ctoys at 
 ditto ? What the Short Practical Rule for pounds, &c. at ditto ? How serve 
 at 5 or 7 per cent. ? What the Short Rule for Federal Moneij, at 6 per cent. T 
 How proceed at 5 or 7 per cent.? What the^r-s^ Rule to compute Interest on 
 Notes, &c. having Endorsements 1 How the Rule contracted ? What the 
 Rule in Massachusetts ? What Compound Interest ? What the Rule 1 What 
 Commission and Brokerage? What Insurance? What Discount? What 
 Present Worth 1 What the Rules? What the /Jw/e when there are several 
 sums to be paid, A:c. ? What an Annuity? How in Arrears? What 
 meant by Amount ? What by Present worth 1 How is the Amount found at 
 Simple Interest ? How the JFVeserrf Worth 1 What Equation of payments ? 
 What the UM/C ? What known by Exchange ? Tell the ToWe. What the 
 Rides? How many &iWs of Vulgar Fractions? What a Proper one? 
 Improper ? Single ? Compound ? Mixed ? How turn a u'/io/e number to a 
 Fraction? What a Complex one? What does a fraction denoted What 
 its value egwa/ to ? What meant by Common Measure 1 W 7 hat by Cornmon 
 Multiple? Give ihe Rules of Pvob\em first. Of Problem second* What is 
 Reduction of Vulgar Fractions ? Repeat the /as* part of the rule in Case 
 first. What the Rule in Case second 1 What in Case third ? In Case fourth ? 
 Those in Case fifth 1 Tell the rule in Case sfri/i ? What the rw/e in Case 
 eighth? W T ha~t are the rule and m?fos in Addition of Vulgar Fractions? 
 What in Subtraction ? What in Multiplication ? What in Division ? How 
 do you proceed in the Rule of Three in Vulgar Fractions*? How in the 
 Rule of Three in Decimals 1 How in the Double Rule of Three hi Vulgar 
 Fractions? Tel! the Table of Ratios in Simple Interest by Decimals. What 
 is Ratiol How do you find the Interest 1 What the Rule in Case second ? 
 In Case third ? Case fourth ? How find Interest for Days ? How calculat- 
 ed on Cash Accounts where partial Payments are made ? How find Com-' 
 pound Interest by Decimals ? How find Amount of an Annuity at Compound 
 Interest 1 How its Present Worth at ditto ? What is Involution 1 What the 
 /rrsf Power ? What the second ? The third 1 The /o?*r/ft ? What Jv0/u- 
 "tt'ori or Extraction of Roots ? What the Root 1 Can the Root of any num- 
 ber be found ? How approximate towards it ? What are Roots called ? 
 What is a Square ? What extracting the Square Root ? What the Rules ? 
 What when a Vulgar Fraction ? What the Rules in its Application and 
 l/*e 7 What a CM/* ? What is, to extract the Cube Root ? What the Rule ? 
 What the note, proposition, and rw/e in its Application and Use ? What iho 
 Rulfs to extract Roots generally 1 What the Note ? When are Numbers in 
 Arithmetical Progression 1 What form increasing ? What decreasing ? 
 How Rawerf ? What TERMS given ? What found I What the whole nura* 
 
TOPICS FOR EXAMINATION. Til 
 
 her called 1 What Problem//^ and rule 1 Second and rule 1 Third and 
 rule ? When are numbers in Geometrical Progression ? What called ratio 1 
 What Problem//^ and rule 1 What Problem second, Case first and rules 1 
 What Case second, rules and note 1 What does Alligation teach ? How 
 distinguished? What Alligation Medial? What the rule! What Alter- 
 natel What the rules 1 What Position 1 * How many kinds 1 What 
 taught by Single Position ? What the rule ? What by Double Position 1 
 What the rules and note 1 What Permutation 1 What Combination ? W hat 
 Problem Jirst and rule 1 Problem second and rule 1 Problem third and rule ? 
 How are G) indstones sold ? How are their Contents found ? What is Su- 
 perficial Measure 1 How made up 1 What is measured by it ? What Case 
 first and rule ? Case second and rule 1 What the Note ? What a Triangle 1 
 *How its Surface measured. How the Superficies of Joists and Planks found ? 
 How measure irregular Surfaces ? What a Circle 1 How find circumfer- 
 ence if diameter be given ? How diameter if circumference be given ? How 
 find Jrea ? How, if circumference a/one be given ? How find diameter by 
 the area ? How circumference by the area ? What is a Sector ? How meas- 
 ured ? By Rule second 1 What a Segment of a Circle ? How find its area 1 
 How measure a regular Polygon 1 How describe an ellipse or oval 1 How 
 find its area ? What is a Sphere or Globe 1 How find its area ? How are 
 solids measured 1 What is a Cube 1 How measured 1 What Case second 
 and rule 1 What noted ? What Case third and rw/e ? What a Cijlinder 1 
 How measured ? What Casejifth and rw/e ? Case sir//i and r?//e? Case 
 
 square pyramid ? MOW, ir a triangular pyramid I Wow, it a circular pyra- 
 mid, or cone ? What is a Globel How find its solid content I What a 
 Frustum of a sphere 1 How find its solid content ? What is Gauging ? 
 What its rw/e and notes ? How use the Sliding rule ? How gauge round 
 tubs ? How a square vessel ? What the Note ? How find a ship's Tonnage ? 
 What the A'o'e? What Section fftk and r?//e? Section sixth and rw/e? 
 How find the solidity of Wboo 1 and Bark ? How find the Cords in a pile of 
 either? What the principal rules in Assessing Taxes? What the general 
 Rules in common; Book-keeping ? 
 
VIII ARITHMETICAL MARKS AND SIGNS. 
 
 ARITHMETICAL MARKS AND SIGNS. 
 
 =The sign of equality and is pronounced, equal to ; 
 +The sign of Addition, and is pronounced, added to ; 
 The sign of Subtraction, and is pronounced, subtract- 
 ed by. 
 
 EXAMPLES. 
 
 12 + 7 = 19, twelve added to seven will be equal to 
 nineteen. 
 
 23 8 = 15, twenty-three subtracted by eight, equal fif- 
 teen. 
 
 xThe sign of Multiplication, and is pronounced, Multi- 
 plied into ; 
 -7-The sign of Division, and is pronounced, divided by. 
 
 EXAMPLES. 
 
 8x7=56, eight multiplied into seven equal fifty-six. 
 36-r-4=9, thirty-six divided by four, equal nine. 
 Division is also implied by the signs 3)6(2 and f =2, 
 six divided by three, equal two. 
 
 : : : : The sign of Proportion, and is pronounced, is to, 
 so is, to. 
 
 EXAMPLE. 
 6 : 9 :: 8 : 12, as 6 is to 9 so is 8 to 12. 
 
 v/ or * signifies the Square Root : thus ^/ 81 is read, the 
 square root of 81 ; or 8l is read 81 in the square root. 
 
 J. 2 
 
 ^3/ or 3 denotes the Cube Root, &c. 3 means that 3 is 
 squared, or to be multiplied, by itself. 
 
 3 4 
 
 3 means that 3 is to be cubed. 48 shows that 48 must 
 be raised to the 4th power. 
 
 19+3x9 = 198 means that 19 added to 3, and the sum 
 multiplied by 9, equal 198 
 
 ~~ 
 
 =3 shows that 12 less the product of 2 multi- 
 
 plied by 3, and divided by 2 equal 3. 
 
ARITHMETIC. 
 
 ARITHMETIC is the art and science of numbers and has 
 for its operation four fundamental rules, viz. Addition, 
 Subtraction, Multiplication, and Division. To under- 
 stand these, it is necessary to have a perfect knowledge 
 of our method of Numeration or Notation. 
 
 NOTATION 
 
 TEACHES to express numbers by words or characters. 
 
 When performed by means of characters or figures, ten 
 are employed. Nine of these are of intrinsic value and 
 are called digits, or significant figures, being written and 
 named thus : 
 
 1 one, 4 four, . 7 seven, 
 
 2 two, 5 five, 8 eight, 
 
 3 three, 6 six, 9 nine. 
 
 The tenth figure, namely, 0, is called naught or cipher, 
 and denotes a want of value wherever it is found. 
 
 Besides the simple value of the digits, as noted above, 
 they have each a local one, which depends on the follow- 
 ing principle. 
 
 In a combination of figures, reckoning from right to 
 left, the figure in the first place represents its simple 
 value ; that in the second place ten times its simple val- 
 ue ; that in the third place an hundred times its simple 
 value ; and so on ; each figure acquiring anew a tenfold 
 value for every higher place it occupies. Hence our sys- 
 tem of arithmetic is called decimal. 
 
 The names of places are denominated according to 
 their order. The first is the place of units ; the second 
 of tens ; the third of hundreds ; the fourth of thousands ; 
 the fifth of ten thousands ; the sixth of hundred thou- 
 sands ; the seventh of millions ; and so on. Thus in the 
 number 8888888 ; 8 in the first place signifies only eight ; 
 8 in the second place eight tens or eighty ; 8 in the third 
 place eight hundred ; 8 in the fourth place eight thousand ; 
 
10 
 
 NOTATION. 
 
 8 in the fifth place eighty thousand ; 8 in the sixth place 
 eight hundred thousand ; 8 in the seventh place eight 
 millions. The whole number is read thus, eight millions, 
 eight hundred and eighty-eight thousand, eight hundred 
 and eighty-eight. 
 
 Though a cipher has no value of itself, yet it occupies 
 a place ; and when set on the right hand of other figures 
 it increases their value in the same tenfold proportion : 
 Thus in the number^SOSO ; the ciphers in the first and 
 third places denote, that, though no simple unit or hun- 
 dreds are reckoned, yet the place of units and that of 
 hundreds are to be kept up to assist in reckoning the 
 tens and thousands. The above number (8080) is read 
 eight thousand and eighty, which, without the two ciphers, 
 would be read eighty-eight. 
 
 Large numbers are divided into periods and half pe- 
 riods, each half period consisting of three figures. The 
 name of the first period is units ; of the second millions ; 
 of the third billions ; of the fourth trillions ; and also, 
 the first part of any period is so many units of it ; and 
 the latter part so many thousands of it.* 
 
 Periods. 
 
 Half do. 
 
 Figures. 
 
 * EXAMPLE. 
 Trillions. Billions. Millions. Units. 
 
 4. 3. 2. 1. 
 
 thou. units. thou. units, thou. units, thou. c. x. u. 
 
 137 462 572 329 484 617 291 337 
 
 g 52 
 
 I I 
 s ' 
 
 > 5 
 
 " 
 
NOTATION. 1 1 
 
 NUMERATION TABLE. 
 
 *>. X xxxxxxxxxxxxxx^xxxx->xxxxxxxxxxxxxxxxxxxxxxx<$- 
 
 v coHundreds of Millions. 
 
 co ooTens of Millions. 
 co GO <iMillions. ^ 
 
 CD QO -5 OiHundreds of Thousands. * 
 CD QO <i a CttTens of Thousands. ^ 
 
 o GO <i o 0* ^Thousands. 
 co OD <? cs CN * csHundreds. 
 \ o QD <! o CK rf* co ioTe-ns. 
 \ cooo<!0:^^coto H-Units. \ 
 
 \ 
 
 To express in figures Numbers which exceed Nine. 
 RULE. Write down ciphers to so many places as are 
 named in the given number ; then, beginning at the left, 
 observe at each place what significant figure is named, 
 and, taking away the cipher, write the significant figure 
 in its place ; and thus proceed with each place till you 
 come to the place of units. 
 
 EXAMPLES. 
 
 Twenty-five, 
 One hundred, 
 
 Three thousand and fifteen, 
 Eight hundred and twelve thousand, 
 Thirty-one thousand, two hundred and six, 
 Six millions, seven thousand and eight, 
 One hundred and one millions, fourteen thousand and 
 fourteen. 
 
 To read NUMBERS. 
 
 RULE. First numerate, from the right to the left hand, 
 each figure, in its proper place, by saying, units, tens^ 
 hundreds, &c., as in the Numeration Table. Then, to 
 the simple value of each figure, join the name of its 
 place, beginning at the left hand, and reading to the right. 
 
 EXAMPLES. 
 64, 
 396, 
 4015, 
 76920, 
 104080, 
 5300648, 
 
SIMPLE ADDITION. 
 
 NOTE. The pupil should be accustomed, in each Example, in the follow- 
 ing Rules, to read correctly not only every answer, but every line of num- 
 bers in his sum. 
 
 ADDITION. 
 
 ADDITION in Arithmetic is the uniting or joining to- 
 gether of two or more numbers. 
 
 SIMPLE ADDITION is the collecting of several numbers, 
 of the same denomination into one sum ; as, 4 yards and 
 6 yards, expressed in one sum, are 10 yards. 
 Addition and Subtraction Table. 
 
 <>XXXXXXXXXXXXXX^XXXXXXXXXXXXX-XXXXX^ 
 
 j 1| 2| 3| 4| 5[ 6| 7| 8| 
 
 2| 4| 5| 6[ 7| 8| 9|10|llj]2|13|14 
 
 \ 3| 5| 6|'7| 8| 9|IOJTTfl2|13|14|15 * 
 
 $"4| 6| 7|"8r9|10jil|ia|13]l4il5fi6^ 
 
 * 5| 7| 8[ 9|10|IT|12|73|14|15|16|17 * 
 } 6[ 8| 9|10|ll|12fl3|14|15|I6|17|18 * 
 
 * 7| 9|10|I1|12|T3|14|I5|16|17|18|19 * 
 
 \ 9|ll|l2|l'3iT4n5[i6|i7|18|19|20|2i \ 
 \ 10|12| I3| 14| 15| I6| 17| 18| I9|20|21 122 ^ 
 
 <^xxxxxxxxxxxxxx^-x~~xxx~xxxx^ 
 
 When you would add two numbers, seek one of them 
 in the left hand column, and the other in the top line ; 
 and in the common angle of meeting, or at the right hand 
 of the first, and under the second, you will find the sum ; 
 as, 6 and 9 are 15; and so of any others. 
 
 When you would subtract, seek, in the left hand col- 
 umn, the number to be subtracted from the greater ; then 
 run your eye along, in the same line, towards the right 
 hand, till you find the number from which the other is to 
 be taken ; and exactly over this last, in the top line, you 
 will find the difference ; as, 6 from 15, and there remain 
 9 ; and so of any others. 
 
 SIMPLE ADDITION. 
 
 RULE. Write the numbers, units under units, tens un- 
 der tens, &c. and draw a line under the whole. Add up 
 the unit column, and if the sum be less than ten, write 
 
SIMPLE ADDITION. 
 
 13 
 
 it under the column ; if it be ten or any number of tens, 
 write a cipher; if there be an excess over ten or tens t 
 write down this excess, and carry as many units to the 
 next column, as there are tens ; and thus proceed with 
 each remaining column, writing the whole sum under 
 the last. 
 
 PROOF. Draw a line below the upper number, and add 
 the remaining numbers as shown in the rule ; add the sum 
 thus found and the upper number together, and if the sum 
 be equal to the first addition, the work is right. Or, 
 begin at the top number, add downwards, and carry as 
 before ; if the two sums come alike, the work is probably 
 right 
 
 EXAMPLES. 
 
 2. 
 
 567842 
 143469 
 782107 
 695213 
 203169 
 
 14786 
 
 4. 
 
 s=3 . o 
 
 S3 3 
 
 98765432 
 46532815 
 90054061 
 00103 
 
 327 
 
 3 2 
 
 15000 
 
 9132051 
 460109 
 
 65400123 
 86194217 
 28103019 
 17631042 
 98765208 
 37849000 
 54001605 
 
 B 
 
14 S13IPLE ADDITION. 
 
 6. 7. 8. 
 
 Miles. Leagues. Years. 
 
 4734746 46434733 347312484 
 
 3474352 74265374 368126312 
 
 4634324 52652754 758612691 
 
 7369138 35374265 731674591 
 
 3543468 74447352 323473276 
 
 4733246 47345264 471266198 
 
 4743447 74167574 323634712 
 
 3752612 43526526 271254712 
 
 7426984 38573452 312844795 
 
 APPLICATION. 
 
 1. What is the sum of 37, 509, 7126, 17630, and 
 459273 yards 1 
 
 2. Required, the sum of 3579, 41, 96120, 725, 11, 
 1820, 5, and 720139 bushels. 
 
 3. What is the sum of 2591, 720396, 14, 259, 6, 
 370214, 9740, 53, 1692, and 137 dollars ? 
 
 4. How many days are in the 12 calendar months, in a 
 leap year ? 
 
 5. A person dying left to his widow 1500 dollars, to his 
 eldest son 30500, to each of his other two 3406 ; also 2700 
 to each of his three daughters, besides 751 dollars in other 
 small legacies ; what did his estate amount to ? 
 
 6. If the distance from Hallowell to Portland be fifty 
 six miles, thence to Portsmouth fifty-four miles, thence to 
 Boston sixty-four Iniles, thence to Hartford ninety-eight 
 miles, thence to New- York one hundred and eleven miles, 
 thence to Philadelphia ninety miles, thence to Baltimore 
 ninety-nine miles, and thence to Washington thirty-eight 
 miles ; what is the whole distance between Hallowell and 
 the city of Washington 1 
 
 7. John, James, and Paul counting their prize-money, 
 John had one thousand, three hundred and seventy-five 
 dollars ; James had just three times as much as John: 
 and Paul had just as much as both the others ; pray how 
 many dollars had Paul ? 
 
SIMPLE SUBTRACTION. 15 
 
 SIMPLE SUBTRACTION. 
 
 SUBTRACTION is finding the difference of two numbers, 
 by taking the less from the greater. It is simple subtrac- 
 tion if the numbers are of one denomination ; as, 5 feet 
 taken from 8 feet, will leave 3 feet. 
 
 The greater number is called the minuend, or substra- 
 tum; the less, the subtrahend; and the number found by 
 the operation, the difference, or remainder. 
 
 RULE. Write the less number under the greater, plac- 
 ing units under units, tens under tens, &c. and draw a 
 line under them. Begin at the right, and take each figure 
 in the subtrahend from its corresponding one in the min- 
 uend, setting down the remainder straight under it below 
 the line. If the lower figure be greater than the one 
 above it, add ten to the upper figure, from which sum take 
 the lower, and set down the remainder, carrying one to 
 the next lower figure ; and thus proceed until the whole 
 is finished. 
 
 PROOF. Add the remainder to the subtraJiend, and if 
 the sum be equal to the minuend, the work is right. 
 
 EXAMPLES. 
 
 1. 2. 3. 
 
 From 67216 the minuend, 46132941 71290 
 
 Take 43792 the subtrahend. 17316257 46172 
 
 23424 Remainder. 
 
 67216 Proof. 
 
 5. 6. * 7. 
 
 87652176 100000 200000 
 
 9107215 65321 99999 
 
 8. 9. 
 
 10000 917144043605 
 
 1 40600S32164 
 
16 SIMPLE MULTIPLICATION. 
 
 10. 11. 
 
 100200300400 1 0000000 
 
 98087076065 9999991 
 
 APPLICATION. 
 
 1. From 360418 tons, take 293752. 
 
 2. From 100046 acres, take 10009. 
 
 3. What is the difference between 1735, and 1897348 
 hours 1 
 
 4. How much do 540312 days exceed 7953 ? 
 
 5. How much are 30491 gallons less than 57321469 ? 
 
 6. If the distance from Hallowell to Savannah, through 
 Washington, be 1268 miles and that from Washington 
 to Savannah, 658 miles ; how far is Washington from 
 Hallowell ? 
 
 7. From Hallowell to the city of New- York is 383 
 miles. Now, if a man should travel 10 days from Hal- 
 lowell towards New-York, at the rate of thirty-six miles 
 each day ; how far would he then be from that city ? 
 
 8. If a farmer kills six hogs, which weigh two hundred 
 and fifty-four, one hundred and ninety-seven, two hun- 
 dred and sixteen, two hundred and forty-nine, three hun- 
 dred and twelve, and three hundred and sixty-three, and 
 markets one thousand weight of pork ; what quantity 
 does he reserve for his own use 1 
 
 SIMPLE MULTIPLICATION. 
 
 MULTIPLICATION is finding the amount of any given 
 number, by repeating it any proposed number of times ; 
 as, 4 times 7 are 28. 
 
 The number to be multiplied is called the multiplicand. 
 
 The number which multiplies is called the multiplier. 
 
 The number arising from the operation is called the 
 product. 
 
 The multiplicand and multiplier are called/actors ; and 
 if these are of one denomination it is called Simple 
 Multiplication. 
 
SIMPLE MULTIPLICATION. 17 
 
 MULTIPLICATION AND DIVISION TABLE. 
 
 } ^V[^^^^\^^\^^\^^\^ \ 
 \ 2| 4| 6| 8|10|12|14|16| 18| 20| 22 1 24$ 
 \ 3| 6| 9|12|15|18J21|24| 27["3QT33| 36 \ 
 * 4 I 8|12|16|20|24|28|32| 36| 4-0 1 44j 48 t 
 $ 5|10|15|20j25|30i35|40|'45| 50| 55 1 60 } 
 $lpp3J24|30|36[42|48| 54|~6Q| 66 1 72* 
 \ 7|14|21|28|35i42|49|56J^3| 70| 77| 84 * 
 x 8|16|24|32i40|48J56|64| 72| 80[ 88 1 96 $ 
 N 9|18i27j36J45{54J63|72| 81| 90 1 99|108 i 
 i 10i20|30[40;50i60|70|80| 9Q|100|UO|12Q | 
 iri|22i^44155|66j77p^9JI^^ ^ 
 
 ^ 12|24|36|48|60 l ,72|84i96ilOSjl20(132|144 $ 
 <^~~ , + +~^ 
 
 USE of the TABLE in MULTIPLICATION. 
 Find the multiplier in the left hand column, and the 
 multiplicand in the uppermost line ; and the product is 
 in the common angle of meeting, or against the multi- 
 plier, and under the multiplicand. 
 
 To use the ahove Table in Division, seek your divisor 
 in the left hand column ; then run your eye along the 
 line, to the right hand, till you come to your dividend ; 
 and the figure in the top line, of the same column, will 
 be the quotient, or number of times the divisor is con-' 
 tain'ed in the dividend. 
 
 CASE I.^When the multiplier is not more than twelve. 
 RULE. Multiply each figure in the multiplicand by the 
 multiplier, beginning at the right hand side, and setting 
 down the whole of such products as are less than ten ; 
 but for such as are just equal to a certain number of tens, 
 write down 0, and carry ] for each ten to the next pro^ 
 duct ; and for such as exceed a certain number of tens, 
 set down the excess, and carry for the tens as before. 
 
 EXAMPLES. 
 
 1. What number is equal to 4 times 365 ? 
 Thus 2. 
 
 365 Multiplicand. 5124167 
 
 4 Multiplier. 3 
 
 Ana. 1460 Product. 
 B3 
 
18 SIMPLE MULTIPLICATION. 
 
 3. 4. 
 
 42179416 74216 
 5 2 
 
 11. 
 
 567295 
 12 
 
 CASE II. When the multiplier consists of several figures. 
 
 RULE. Set the multiplier under the multiplicand, so 
 that units may be under units, tens under tens, <fec. then 
 find the product for each figure in it, as in the first case t 
 not regarding in what order the lines are found, provided 
 the first figure in each stand straight below its respective 
 multiplier. Add all the lines of products together in the 
 same order as they stand, and the sum will be the whole 
 product required. 
 
 PROOF. Make the former multiplicand the multiplier, 
 and the multiplier the multiplicand, and proceed as be- 
 fore ; and the new product will be the same as before, 
 when the work in both is right. Or, add together the fig* 
 ures first of one factor, and then of the other, casting 
 out all the nines in the sums of each, as often as they 
 amount to 9. Multiply the two remainders, if any, to- 
 gether, and the nines cast out of their product, will leave 
 the same remainder as the nines cast out of the answer, 
 when the work is right. The first remainder may be set 
 at the left side of the cross, or X ; the second at the right ; 
 that arising from their product at the top ; and that aris- 
 ing from the answer at the bottom ; if the answer ba 
 right, the top and bottom figures will be alike* 
 
SIMPLE MULTIPLICATION. 19 
 
 EXAMPLES. 
 
 I. 329 excess of 9s, 5 
 
 4271 Multiplicand. 4271 do. do. 5 
 329 Multiplier. 
 
 329 
 
 38439 2303 
 
 8542 658 
 
 12813 1316 
 
 1405159 Product. Proof 1405159 excess of 9s, 7 
 
 7 
 or 5x5 
 
 7 
 
 2. 3. 4. 
 
 691861 129186 281216 
 
 26 98 978 
 
 17988386 12660228 275029248 
 
 5. 6. 7. 
 
 181281 281691 264648436 
 763 76286 3639604 
 
 138317403 21489079626 963215506259344 
 
 CONTRACTIONS IN MULTIPLICATION. 
 
 CASE I. When there are ciphers at the, right hand of one 
 
 or both of the factors. 
 
 RULE. Proceed as before, neglecting the ciphers, and 
 to the right hand of the product, place as many ciphers 
 as are in both the numbers. 
 
 EXAMPLES. 
 
 1. 2. 3. 
 
 27600 180120 27640 
 
 48000 48100 20 
 
 2208 
 1104 
 
 1 324800000 8663772000 
 
20 SIMPLE MULTIPLICATION. 
 
 CASE II. Wfien the multiplier is the product of two or 
 
 more numbers. 
 
 RULE. Multiply once successively by each of those 
 numbers instead of using the whole multiplier at once, 
 
 EXAMPLES. 
 
 1. % 
 
 Multiply 7629 by 63 74639 by 72 
 
 7x9=63 7 
 
 3 
 46217 by 96 
 
 4. 
 
 Prod. 480627 37692 by 132 
 
 APPLICATION. 
 
 1. What will 37 horses for shipping come to, at 52 
 dollars per head 1 Ans. 1924 dols. 
 
 2. What will 587 firkins of butter come to, at 7 dol- 
 lars per firkin ? Ans. 4109 dols. 
 
 3. What will 367 acres of land cost, at 13 dollars per 
 acre ? Ans. 4771 dols. 
 
 4. If a barrel of pork cost 18 dollars what will 857 
 barrels be worth ? Ans. 15426 dols. 
 
 5. What will be the worth of 924 tons of potash, if 
 one ton sell for 95 dollars 1 Ans. 87780 dols. 
 
 6. A merchant having traded ten years, found he was 
 worth 13000 pounds. His books showed that the last 
 three years he had cleared 873 pounds a year ; the three 
 preceding but 586 pounds a year ; and before that but 
 364 pounds a year. With what sum did he begin busi- 
 ness ? Ans. 7167 pounds. 
 
 7. Trajan's bridge over the Danube is said to have had 
 twenty piers to support the arches, every pier being 60 
 feet thick, and some of them .150 feet above the bed of 
 the river ; they were also 170 feet asunder. Pray, how 
 wide was the river in that place 1 and how much did this 
 bridge exceed in length that at Westminster, in England, 
 which is about 1200 feet from shore to shore, and is sup- 
 ported by 11 piers, making the number of arches 12 I 
 
 i f 4770 feet wide, and 3570 feet lon~ 
 5 * ( ger than Westminster bridge, 
 
SIMPLE DIVISION. 21 
 
 8 In the partition of lands in a certain settlement, A. 
 had 757 acres allotted to him ; B. 2104 ; C. 16410 ; D. 
 12881; E. 11008; F. 9813; H. 13800; and I. 8818 
 acres. Now as the above allotments want 416 acres to 
 make them just one fifth of the whole, how many acres 
 did the settlement contain T Ans. 380035 acres. 
 
 DIVISION. 
 
 DIVISION shows how often one number is contained in 
 another : as 24 divided by 6, produce 4 in the quotient ; 
 that is, 6 are contained 4 times in 24. 
 
 The number to be divided is called the dividend. 
 
 The number by which we divide is called the divisor. 
 
 The number of times the dividend contains the divisor 
 is called the quotient. 
 
 The remainder, if there be any, will be less than the 
 divisor. 
 
 It is called Simple Division, if the dividend and divisor 
 have but one and the like name. 
 
 RULE. On the right and left of the dividend draw a 
 curved line, and write the divisor on the left, and the quo- 
 tient as it arises on the right hand. Assume as many fig- 
 ures on the left hand of the dividend as contain the divi- 
 sor once, or more, and place the number in the quotient. 
 Multiply the divisor by the quotient figure, and set the 
 product under the assumed part of the dividend ; subtract 
 it, and to the remainder bring down the next figure of the 
 dividend ; which number divide as before, and thus pro- 
 ceed until the whole is divided. 
 
 NOTE 1. If after a figure is brought down, the number 
 be less than the divisor, place a cipher, in the quotient, and 
 bring down the next figure of the dividend. 
 
 NOTE 2. Remember that the products of the divisor 
 and the several quotient figures, must always be less than 
 the parts of the dividend under which they are set, unless 
 they chance to be just the same numbers ; and that every 
 remainder must be less than the divisor. 
 
 PROOF. To the product of the divisor and quotient, 
 add the remainder, which sum will be equal to the divi- 
 dend, if the work is right. 
 
SIMPLE DIVISION. 
 
 Or, cast the nines out of the divisor, and place the ex- 
 cess or remainder on the leftside of a cross, or X ; do the 
 same with the quotient, and place the excess on the right 
 hand ; multiply these two figures together, add their pro- 
 duct to the remainder of the divisor, if there be any, cast 
 out the nines in the sum, and set the excess at the top of 
 the cross; cast the nines Out of the dividend, and place 
 the excess at the bottom ; then, if the top^ and bottom 
 figures are alike, the work is right. 
 
 EXAMPLES. 
 I 2 
 
 Divis. Divid. Quot. 
 48)7641312(159194 
 
 48 48 ' 15 
 
 Divis. Divid. Quot. 
 5)172164(34432f 
 
 284 
 
 1273552 
 
 240 636776 
 
 441 
 432 
 
 7641312 Proof. 
 
 22 
 20 
 
 21 
 20 
 
 93 Proof by excess 
 
 48 of nines. 
 
 6 
 
 451 3x2 
 
 432 6 
 
 192 
 192 
 
 16 
 15 
 
 14 
 10 
 
 Proof. 
 
 3 
 
 5x7 
 3 
 
 5x7=35 
 add 4 rem. 
 
 39 
 
 4 Remainder. 
 
 NOTE. If there be no remainder, the quotient is the 
 perfect answer to the question ; but if there is, to com- 
 plete the quotient, put the remainder at the end of it, and 
 the divisor below it, drawing a line between the two. 
 
 Quotient. Rem. 
 by 29 Ans. 5296 and 14 
 63 4780 7 
 
 763 181281 
 
 232 48340 7 
 
 400 11695 216 
 
 3215 
 
 2359 1255 
 
 6000 4711 
 
 3. Divide 153593 
 
 4. 
 
 301147 
 
 5. 
 
 138317403 
 
 6. 
 
 11214887 
 
 7. 
 
 4678216 
 
 8. 
 
 1030603615 
 
 9. 
 
 4917968967 
 
 10, 
 
 210634711 
 
SIMPLE DIVISION. 23 
 
 CONTRACTIONS. 
 
 CASE I. When there are ciphers at the right hand of 
 the divisor, cut them off; likewise cut off the same num- 
 ber of digits from the right hand of the dividend ; then 
 divide as usual, and to the remainder annex the digits 
 cut off from the dividend. 
 
 EXAMPLES. 
 
 1. 2. 
 
 342,00)6792,16(19 135,000)27619,413( 
 
 342 
 
 3372 
 3078 
 
 29416 Remainder. 79413 Rem. 
 
 CASE II. When the divisor is any number not exceed- 
 ing 12. 
 
 RULE. First seek how often the divisor can be had in 
 the first figure, or figures, of the dividend ; put the result 
 under the dividend ; multiply this quotient figure and the 
 divisor together ; mentally subtract their product from, the 
 part of the dividend taken ; what remains call so many 
 tens, which place, in idea, before the next figure of the 
 dividend for a new dividual ; and so proceed through the' 
 whole dividend. When in subtracting, nothing remains, 
 take the next figure ; if that be less than the divisor, 
 take the next two, and place a cipher under the first. 
 
 1. 2. 
 
 6)7241324( 5)172164( 
 
 Quotient 1206887f Rem. Quotient 34432f Rem. 
 
 3. Divide 3764592 by 7 
 
 4. 527684 8 
 
 5. 1410217 9 
 
 6. 612948 11 
 
 7. 317926 12 
 
24 SIMPLE DIVISION. 
 
 CASE III. If the divisor be a product of two or more 
 numbers, divide continually by those numbers instead of 
 the whole at once. 
 
 EXAMPLES. 
 
 1. 2. 
 
 Divide 7621460 by 16 4792161 by 48 
 
 4)7621460 ' 6)4792161 
 
 4)1905365 8) 3 
 
 Quo. 4763414 Rem. 998365x6+3=33 
 
 NOTE. It sometimes happens that there is a remainder 
 to each of the quotients, and neither of them the true one 
 but the true remainder maybe found by the following rule. 
 
 RULE. Multiply the last remainder by the last divisor 
 but one, and to the product add the preceding remainder ; 
 multiply this sum by the next preceding divisor, and to 
 this product add the next preceding remainder, and so 
 on until all the remainders and divisors are used ; and 
 the last sum will be the true remainder. 
 
 3. 4. 
 
 Divide 6421671 by 448 Divide 27162 by 62 
 8x8x7=448 8)6421671 
 
 8)8027087 
 
 7)1003384 
 Quotient. 143344x8+7=39 Remainder. 
 
 APPLICATION. 
 
 1. Divide 3656 dollars equally among 8 men. 
 
 Ans. 457 dols. to each. 
 
 2. There are 124 men who have 372 dollars among 
 them ; how much is one man's share, if it be divided 
 equally ? Ans. 3 dols. 
 
 3. If I wish to perform a journey of 3264 miles in 136 
 days, how far must I travel each day to complete it ? 
 
 Ans. 24 miles* 
 
PRACTICAL 'QUESTIONS. 25 
 
 4. A payment of 1272 dollars was made by a number 
 of men, each of whom paid 3 dollars ; how many men 
 were there T Ans. 424. 
 
 5. I would plant 2072 trees, in 14 rows, 25 feet asun- 
 der; how long must the grove be . ? Ans. 3675 feet. 
 
 6. Divide 1000 dollars, between A, B, and C, and give 
 A 129 more than B, and B 178 less than C. 
 
 Ans. 360 dots. A's, 231 B's, and 409 C's. 
 
 7. Part 1500 acres of land between Saul, Seth, and 
 Silas; and give Seth 72 more than Saul, and Silas 112 
 more than Seth. . ( 414f Saul's share, 486f 
 
 3 ' \ Seth's and 598f Silas's. 
 
 8. A brigade of horse consisting of 384 men, is to be 
 formed into a column, having 32 men in front ; how 
 many ranks will there be ? Ans. 12. 
 
 9. In order to raise a joint stock of 10,000 dols. L, M, 
 and N, together, subscribe 8500, and O the rest. Now, 
 M and N are known together to have set their hands to 
 6050, and N has been heard to say that he had undertaken 
 for 420 more than M. What did each proprietor advance? 
 
 Ans. L 2450, M 2815, N 3235, and O 1500. 
 
 PRACTICAL QUESTIONS, 
 
 UNDER THE PRECEDING RULES. 
 
 1. Add fourteen thousand, five hundred and nine ; one 
 thousand, nine hundred and twenty-one ; six hundred and 
 twenty thousand, three hundred and forty-seven; and live 
 million, twenty-three thousand, arid nineteen, together. 
 
 Ans. 5659796 sum. 
 
 2. What is the sum of 76129+54216+39127+62357 
 + 514026? Ans. 745855. 
 
 3. What is the difference between four million two 
 hundred and ten thousand and twelve; and six hundred 
 and fifty-nine thousand seven hundred and ninety-seven 1 
 
 Ans. 3550215. 
 
 4. Take nine hundred and one thousand and fifteen, 
 from one million one thousand one hundred and one ? 
 
 Ans. 100086. 
 
 5. A farm of 460 acres is let for 2 dollars per acre ; 
 bow much does the rent amount to ? Ans. 920 dols. 
 
 C 
 
26 PRACTICAL QUESTIONS. 
 
 6. If a man's income be 6 dollars a day, how much 
 does it amount to in a year, allowing 365 days in a year ? 
 
 Ans. 2190 dollars. 
 
 7. What is the product of 376x54 ? Ans. 20304. 
 
 8. 64 men have 17280 dollars divided equally among 
 them ; what is each man's part ? Ans. 270 dollars. 
 
 9. Multiply three hundred and seventy-eight thousand 
 and five hundred, by thirty-four. Ans. 12869000 product. 
 
 10. What is the third part of 3669 1 Ans. 1223. 
 
 11. Divide 6764 by 19. Ans. 356 quotient. 
 
 12. What number must be added to 764 to make it 
 1256 ? Ans. 492. 
 
 13. By what number must I multiply 67, that the pro- 
 duct may be 871 ? Ans. 13. 
 
 14. There are two numbers whose difference is 796, 
 the greater number is 4320 ; I demand the less. 
 
 Ans. 3524. 
 
 , 15. Supposing a man to have been born in the year 
 1762 ; how old was he in 1806 ? Ans. 44. 
 
 16. Suppose a man to have been 78 years old in the 
 year 1806 ; in what year was he born ? Ans. 1728. 
 
 17. What will 12 tons of hay come to at 27 dollars 
 per ton ? Ans. 324 dollars. 
 
 18. What will 750 barrels of beef come to at 11 dol- 
 lars per barrel ; and what will the profits amount to in 
 selling it, if I clear 3 dollars on each barrel ? 
 
 A ( 8250 dollars amount. 
 '* \ 2250 dollars profit. 
 
 19. There is a town which contains 290 houses, and 
 each house 6 inhabitants ; how many inhabitants are 
 there in that town ? Ans. 1740. 
 
 20. A prize of 48726 dollars is owned by 270 men ; 
 what is each man's share ? Ans. ISOf-J dollars. 
 
 21. If 12 bundles of wheat produce I bushel, how many 
 bushels will 4764 bundles produce ? Ans. 397 bushels. 
 
 22. Borrowed of AJ 12 sums of money, each 250 dol- 
 lars ; paid him at one time 97 dollars, and at another 35 ; 
 the balance 1 am to pay him in six equal payments ; what 
 is one of those payments 7 Ans. 478 dollars. 
 
MONEY, WEIGHTS, MEASURES, &C. 2? 
 
 TABLES 
 
 OF MONEY, WEIGHTS, AND MEASURES. 
 1. MONEY,* 
 
 4 Farthings make one penny ; qr. d. denote farthings 
 
 and pence respectively. 
 
 12 Pence make one shilling - - 5 - - Shilling. 
 20 Shillings 1 pound .--,-- Pound. 
 Is one farthing, or one fourth ; % is one halfpenny, 
 or one half; three farthings, or three fourths. 
 
 2. TROY WEIGHT. 
 
 24 Grains make one pennyweight, marked grs. dwt. 
 20 Pennyweights - 1 Ounce, - - - oz. 
 
 12 Ounces - - 1 Pound, - - fo- or Ib. 
 
 By this weight are weighed jewels, gold, silver, electu- 
 aries and liquors, 
 
 3. APOTHECARIES' WEIGHT. 
 
 80 Grains make - 1 Scruple, marked gr. J),orser. 
 3 Scruples - - 1 Dram, - - - 3, or rfr. 
 8 Drarns - - - 1 Ounce, - - - Z, or oz. 
 12 Ounces - - - 1 Pound, - - - - jfc. or Ib. 
 Apothecaries use this weight in compounding their 
 medicines, but they buy and sell their drugs by Avoirdu- 
 pois weight. 
 
 * Sterling- money was, fornrafy, of the same value in all the 
 Colonies of North- America. y reason, however, of the emis- 
 sion of paper money by the Legislatures of those Colonies, 
 which afterwards depreciated, the Spanish dollar came to be 
 reckoned, in different Colonies, at a higher or lower value, ac- 
 cordingly to the less or greater depreciation of their paper cur- 
 rencies. Still, though the pound was valued accordingly to this 
 paper medium, it was, in every Colony, reckoned at twenty shil- 
 lings, as in England. Thus, a Spanish dollar being- worth 4*. 6d. 
 in England, became, in Georgia and South- Carol in a, where the 
 depreciation of the paper was least, worth 4s. 8d. ; in Canada 
 and Nova-Scotia, where it was somewhat greater, 5s. ; in New- 
 England, Virginia, Kentucky, and Tennessee, 6s ; in New- Jer- 
 sey, Pennsylvania, Delaware and Maryland, 7s. 6d. ; and in 
 New-York, and North-Carolina, 8s. 
 
WEIGHTS, MEASURES, &C. 
 
 4. AVOIRDUPOIS WEIGHT. 
 
 16 Drams make 
 16 Ounces 
 28 Pounds* 
 
 4 Quarters - 
 20 Hundred wt. 
 
 1 Ounce, marked, - dr. oz. 
 1 Pound, - Jk or Tb. 
 
 1 Quarter, - - - qr. 
 1 Hundred weight. - cwt. 
 1 Ton, ... T. 
 
 By a late law of this State, Soffomake a qr. 
 
 By this weight are weighed all things of a coarse na- 
 ture ; such as leather, cheese, grocery wares, bread, and 
 all metals except gold and silver. It is our common 
 steelyard weight. t 
 
 NOTE. 6760 grains 1ft. Troy; 7000 grains 1ft. 
 Avoirdupois ; therefore the weight of a pound Troy, is to 
 a pound Avoirdupois as 5760 to 7000, or as 144 to 175. 
 
 5. CLOTH MEASURE. 
 
 4 Nails, or 9 inches, make 1 Quarter, marked, na. qr, 
 
 4 Quarters - - 1 Yard, - yd 
 3 Quarters - - - 1 Ell Flemish, - - E. Fl 
 
 5 Quarters - - - 1 Ell English, - E. E 
 
 6 Quarters - 1 Ell French, - - E. Fr\ 
 
 NOTE. We buy Scotch and Irish linens by our Amer- 
 ican yard, and Dutch linens by the eli Flemish ; but we 
 sell them here by the same measure, the yard. 
 
 * Most of ^e merchants and traders in the United States, now 
 call 25 Ibs. onlyj^ quarter of a cwt. 
 
 f A Firkin of Butter is 56 Ib. 
 
 A firkin of Soap 64 
 
 A Barrel of Beef 220 
 
 Pork 220 
 
 Potashes 400 
 
 Soap 256 
 
 Butter 224 
 
 Flour 196 
 
 Gunpowder 112 
 
 Raisins 112 
 Fish 30 gallons 
 
 A Quintal of Fish 112lb. 
 
 of Iron 14 
 
 n of Train Oil 7 
 
 20 Thing-s make 1 Score. 
 
 12 do. 1 Dozen. 
 
 12 Dozen 1 Gross. 
 
 144 do. 1 Great Gross. 
 
 A Quire of Paper 24 Sheets. 
 
 A Ream of Paper 20 Quires. 
 
 A Bale of Paper 10 Reams. 
 
 A Roll of Parchment 60 Skins. 
 
 Hoops and Staves are now reckoned five scores to the hun- 
 dred in this State, by a late law. 
 A ton in weight for Ships is 2000 Ib. 
 A ton for goods, boxes, cases, &c. is 40 cubic feet. 
 
MEASURES. 29 
 
 6. LONG MEASURE. 
 
 3 Barley Corns make 1 Inch, marked Bar. In. 
 
 12 Inches 1 Foot, - Ft. 
 
 3 Feet, - - 1 Yard, - - Yd. 
 
 5 Yards, or 16 Feet 1 Pole, Rod or Perch, Rod. 
 
 40 Poles or Rods 1 Furlong, - - - Fur. 
 
 8 Furlongs - - - 1 Mile, - Mi. 
 
 3 Miles - 1 League, - Lea. 
 60 Geographical, or \ * DpOTPP rw 
 
 69 Statute Miles } 1>e ^ ree ' JJe ^' 
 
 360 Degrees make the circumference of the Earth. 
 
 By this measure distances are measured. 
 66 feet, or 4 rods, make a Gunter's chain, containing 
 
 100, links, each of which is 7 T 9 /o inches. 
 6 feet make a fathom, in measuring depths. 
 5 feet make a geometrical pace. 
 
 4 inches make a hand, in measuring the height of 
 horses. 6ft. 44in=a French toise. 
 
 1 French post=% Fr. Leagues=5 T 5 ^ Eng. miles. 
 1 German short mile =3^^^ Eng. miles. 
 1 Eng. mile = lJ+Russian verst. 
 
 7. LAND OR SQUARE MEASURE. 
 
 144 Square inches make 1 Square Foot, marked In.Ft. 
 9 Feet - - 1 Yard, - - Yd. 
 
 272* Feet* } " ' l Rod ' Pole r Percb ' Rod ' 
 
 40 Rods 1 Rood or ^ of an acre, Rood. 
 
 4 Roods - - 1 Acre, - - Ac. 
 
 640 Acres 1 Mile, - - Mi. 
 
 By this measure, surfaces* are measured. It is long 
 
 measure squared, or multiplied into itself. 
 
 8. CUBIC OR SOLID MEASURE. 
 
 1728 Solid Inches make 1 Foot, Marked In* Ft. 
 
 27 Feet 1 Yard, Yd. 
 
 50 Feet of hewn, or ) t m T j TT 
 
 40 Feet of round Timber } l Ton or Load ' 
 128 Feet, i.e. 8 feet in length ) c d f w d c 
 
 4 in breadth and 4 in height ) 
 
 By this measure the contents of solids are obtained, 
 or things that have length, breadth, and depth. It is long 
 measure cubed, or multiplied by itself, twice, 
 C 2 
 
30 MEASURES. 
 
 9. DRY 
 
 MEASURE. 
 
 2 Pints make - - - 
 
 1 Quart, marked 
 
 4 Quarts 
 
 1 Gallon, 
 
 2 Gallons 
 
 1 Peck, - 
 
 4 Pecks - 
 
 1 Bushel, - 
 
 8 Bushels 
 
 1 Quarter, 
 
 36 Bushels - 
 
 1 Chaldron, - 
 
 Pt. Qt. 
 
 Gal. 
 
 Pk. 
 Bus. 
 
 Qr. 
 
 dial 
 
 8 bushels a Hogshead of Salt. 
 
 NOTE. The diameter of ihe Winchester or common 
 bushel is 18| inches, and its depth 8 inches. 
 
 The gallon dry measure contains 268f cubic inches. 
 
 Corn, grain, beans, peas, flax-seed, salt, coals, &c. are 
 measurecf by striked measure ; but pears, apples, turnips, 
 potatoes, onions, &c. are heaped to a handsome round- 
 ing measure. The bushel contains 2150f + cubic inches. 
 
 10. WINE MEASURE. 
 
 4 Gills make - 1 Pint, marked - Gill Pt. 
 
 2 Pints 1 Quart, Qt. 
 
 4 Quarts - I Gallon, - Gal. 
 
 42 Gallons - 1 Tierce, - Tier. 
 
 63 Gallons - 1 Hogshead, - Hhd. 
 
 84 Gallons - 1 Puncheon, Punch. 
 
 2 Hogsheads 1 Pipe or Butt, - Pipe. 
 
 2 Pipes or 4 Hhds. - 1 Tun, Tun. 
 
 NOTE. The wine gallon contains 231 cubic inches. 
 The hogshead of 63 gallons, and the puncheon of 84 
 gallons, are not used with us. The hogshead of 108 or 110 
 gallons is called a hogshead or a puncheon. Brandies, spir- 
 its, perry, cider, vinegar, mead, oil and honey, are sold by 
 this measure, though honey is|gmetimes sold by the pound 
 avoirdupois. Milk is sometimes measured by this meas- 
 ure, though more commonly and justly by beer measure. 
 11. ALE MEASURE. 
 
 2 Pints make 1 Quart, marked - Pt. Qt. 
 
 4 Quarts - 1 Gallon, Gal. 
 
 5 Gallons 1 Firkin of Ale, - A. Fir. 
 
 9 Gallons - 1 Firkin of Beer - B. Fir. 
 36 Gallons 1 Barrel of Beer, Bar. 
 54 Gallons - 1 Hogshead, Hhd. 
 
 3 Barrels 1 Butt, - Butt. 
 NOTE. The Ale gallon contains 282 cubic inches. 
 Milk is sold by the Beer quart, which is about one sixth 
 
 larger than the wine, cider, &c. quart. 32 gal.= Ibar. ale* 
 
TIME, MOTION. 31 
 
 12. TIME. 
 
 60 Seconds make 1 Minute, marked 8. M. 
 
 60 Minutes 1 Hour, H. 
 
 24 Hours 1 Day, - D. 
 
 7 Days 1 Week, W. 
 
 4 Weeks - : 1 Month, - - Mo. 
 
 13 Lunar or 12 Solar months ) t v v 
 
 nr>r TV i * JL CclT, - JL 
 
 or 365 Days J 
 
 NOTE. 305 days, 5 hours, 48 minutes, 48 seconds, 
 make a solar year according to the most exact observation. 
 
 April, June, September and November, have each 30 
 days ; each of the other months has 31, except February r 
 which has 28 in common years and 29 in leap years.* 
 
 30 years make an age, and 100 years a century, 
 
 A lunar month is 29d. 12h. 44m. 3s. nearly. 
 
 13. CIRCULAR MOTION. 
 
 60 Seconds make - 1 Prime minute, marked " ' 
 60 Minutes - 1 Degree, 
 
 30 Degrees - - 1 Sign, - & 
 
 12 Signs, or ) ( The whole circle 
 
 360 Degrees J (of the Zodic. 
 
 * To find whether any given year will be leap year. 
 RULE. Divide the given year by 4 ; if nothing- remain, it is 
 leap year ; but if there be a remainder, that is the number of 
 years after leap year. 
 
 EXAMPLE. 
 Was 1823 leap year ? 4)1823 
 
 455-3 rem. 
 which shows it to have been the 3d after leap year. 
 
 The last year in every three centuries out of four, which would 
 otherwise be leap year, is to be reduced to a common year. 
 
 To find whether the last year in any given century is leap year. 
 
 RULE. Divide the given century only, or the hundreds in the 
 year, by 4 ; if nothing- remain, it is leap year ; but a remainder 
 shows it is to be counted a common year. 
 
 EXAMPLE. 
 
 Will the year 1900 be leap year ? 
 4)19(4 
 16 
 
 3 remainder ; therefore a common year. 
 
82 REDUCTION. 
 
 REDUCTION. 
 
 REDUCTION teaches to change the denomination of 
 numbers without altering their value. 
 
 RULE. Wfcen the reduction is from a higher denomi- 
 nation, to a lower, as pounds into shillings, tons into oun- 
 ces, &c. multiply the highest denomination by as many 
 of the next lower as make one of the highest, adding to 
 the product the parts of the same name ; multiply this 
 sum by the next lower, adding to the product the parts of 
 its own name, if any ; and so on to the denomination 
 required. 
 
 When the reduction is from a lower to a higher denomi- 
 nation, pence into pounds, minutes into days, &c. divide 
 the given number by as many of that denomination as 
 make one of the next higher, and so on, to the denomi- 
 nation required ; and the last quotient with the several 
 remainders, (if any) will be the answer required. 
 
 The proof is had by reversing the question. 
 
 EXAMPLE. 
 MONEY. 
 
 1. In 476 pounds, how many shillings and pence 1 
 476 
 20 
 
 9520 Shillings. 
 
 12 12)114240 
 
 Ans. 114240 Pence. 2,0)952,0 
 
 Proof, 476 
 
 2. In 3694 shillings, how many pence ? Ans. 44328. 
 
 3. How many farthings in 69217 pence 1 Ans. 276868. 
 
 4. Reduce 6942 pounds to farthings. Ans. 6664320. 
 
REDUCTION. 33 
 
 5. In c49 195. llfd. how many farthings? 
 
 s. d. qr. 
 49 19 11 3 
 20 
 
 999 Shillings, 
 12 
 
 11999 Pence. 
 4 
 
 Ans. 47999 Farthings. 
 
 6. How many Pence in c472 13s. 4d. ? 
 
 Ans. 113440, 
 
 7. How many pounds in 46*216 farthings ? 
 
 4)467216 
 
 12)116804 
 2,0)973,3 Sd. 
 
 Ans. ,486 13s. Sd. 
 
 8. How many pounds in 9752 pence 1 
 
 Ans. <40 12s. Sd. 
 
 9. In 648 English guineas,* how many pence ?. 
 
 Ans. 217728. 
 * TABLE, 
 
 Showing the weight and value of several pieces of For- 
 eign Coins. 
 
 s. d. $ cts. 
 
 An English Shilling is - - 1 4 or 22-f 
 
 A French Franc, - 1 1 18$ 
 
 A Livre Tourriois, - . 1 ]J 18J 
 
 An English or French Crown, - 6 7.1 110 
 
 Napoleon, 4 dwt. 6 grs. - 123 3 71 
 
 4 Johannes* 9 - - 280 8 00 
 
 Moidore, 6 18 1 16 6 00 
 
 Eng. Guinea, 5 6 - 1 8 4 662 
 
 French do. 5 5 - 1 7 3i 4 54 
 
 Sp. Pistole, 4 5 - 113 3 54^ 
 
 By an Act of Congress passed April 29th, 1816, the gold coins 
 of Great Britain and Portugal, are estimated at 27 grains to the 
 dollar; those of France at 27^ grains to the dollar; and those 
 of Spain and her dominions, at 28^ grains to the dollar. 
 
34 REDUCTION. 
 
 10. How many Eng. guineas in 28560 pence ? Ans. 85. 
 
 11. In 37 Spanish pistoles, how many farthings? 
 
 Ans. 37740. 
 
 12. In 48960 farthings, how many pence, three-pences, 
 six-pences and dollars 1 Ans. 12240 pence, 4080 three- 
 
 pences, 2040 six-pences and 170 dollars. 
 
 13. In 427 moidores, how many dollars and pounds ? 
 
 Ans. $2562, or ,768 12s. 
 
 14. In 11040 pence, how many dollars? Ans. $15325. 
 
 15. How many pounds in 91751 farthings ? 
 
 Ans. .95 11s. 
 
 TROY WEIGHT. 
 
 1. How many grains in 47ft. 100z. of gold ? 
 
 Ans. 275520. 
 
 2. In 47128 grains of gold, how many pounds ? 
 
 Ans. 8ft. 20z. Sdwt. 16grs. 
 
 3. In 5605 grains, how many ounces ? 
 
 Ans. lloz. \3dwt. ISgrs. 
 
 4. How many grains in 18ft. 5oz. 9dwt. 21grs. ? 
 
 Ans. 
 
 AyontBtfpots WEIGHT* 
 
 1. In }%cwt Iqr. 18ft. how many ounces, at 25ft. to 
 the qr. of a cwt ? Ans. 19888. 
 
 2. How many tons in 3440640 drams ? Ans. 6 Tons. 
 
 3. In27 f . \5cwt. 2grs. 17ft. how many pounds, at 
 25ft. to the quarter ? Ans. 5567ft* 
 
 4. How many pieces of 4ft. 5jft. and 6jfc. of each 
 an equal number, in 6%cwt. %qrs. 24ft. of beef? 
 
 Ans. 439 pieces of each. 
 
 5. In 3 tons of hay, how many pounds ? Ans. 6720ft. 
 
 APOTHECARIES' WEIGHT. 
 
 1. In 31ft. 2^. 65. how many drams. Ans. 2998. 
 
 2. How many pounds in 2535 scruples 1 
 
 Ans. Sft. 9. 53. 
 
 CLOTH MEASURE. 
 
 1. How many quarters in 83yds. Sqrs.1 Ans. 335. 
 
 2. In 2528 nails, how many yards ? Ans. 15. 
 
 3. In 840 nails, how many ells English ? Ans. 42. 
 
REDUCTION. 35 
 
 4. How many yards of Holland in 58 pieces, each con- 
 taining 36 ells Flemish 1 Ans. 1566. 
 
 5. In 748 ells French, how many ells English, ells 
 Flemish, yards, quarters, and nails 1 Ans. 897 E. E. 
 3 ^rs. 1496 E. Fl 1122 yds. 4488 qrs. 17952 na. 
 
 LONG MEASURE. 
 
 1. In 70 miles, how many furlongs and poles ? 
 
 Ans. 560 fur. 22400 poles. 
 
 2. How many leagues in 21120 yards 7 Ans. 4. 
 
 3. How many barley corns in 360 degrees, each degree 
 69 miles ? Ans. 4755801600. 
 
 4. How often will a wheel that is 15 feet in circumfer- 
 ence, turn round in the distance from Hallowell to Farm- 
 ington, it being 32 miles 1 Ans. 11264. 
 
 LAND OR SQUARE MEASURE. 
 
 1. In 17 acres 3 roods 10 poles, how many poles ? 
 
 Ans. 2850. 
 
 2. In 815443200 inches, how many acres 1 Ans. 130. 
 
 3. How many acres in 6654 rods or poles ? 
 
 Ans. 41 Acres, 2 Roods, 14 Rods. 
 
 4. In 6 acres 1 rood, how many perches 1 Ans. 1000. 
 
 5. If a room be 16 feet long, and 14 feet wide, how 
 many feet of boards will it take to lay the floor 1 
 
 Ans 224 feet. 
 
 6. How many shingles will it take to cover the roof of 
 a house 40 feet in length, and of 18 feet rafters, allowing 
 each shingle to be 4 inches wide, and each course to be 
 laid out 6 inches ? Ans. 8640. 
 
 7. How many boards will cover a barn that is 50 feet 
 long, and 30 feet wide ; the height of the gable ends 13 
 feet, and the rafters 20 feet each ; and the posts, or body 
 of the frame, 15 feet in height ? Ans. 4790 feet. 
 
 CUBIC OR SOLID MEASURE. 
 
 1. In 9 tons of round timber, how many inches ? 
 
 Ans. 622080. 
 
 2. How many cords of wood in 3096576 inches 1 
 
 Ans. 14. 
 
 3. In 259200 inches of hewn timber, how many tons ? 
 
 Ans. 3. 
 
 4. How many bricks, 8 inches long, 4 wide, and 2 
 thick, will build a house 44 feet long, 40 feet wide, and 20 
 feet high, with walls 12 inches thick ? Ans. 88560. 
 
36 
 
 REDUCTION. 
 
 DRY MEASURE. 
 
 1. In 49 bushels, how many quarts ? Ans. 1568. 
 
 2. How many bushels in -27072 quarts ? Ans. 846. 
 
 3. How many pints in 150 bushels of. corn ? Ans. 9600. 
 
 4. In 56 bushels of wheat, Canada measure, how 
 many bushels of the United States 1 Ans. 70. 
 
 NOTE. 5 pecks, or 40 quarts, make 1 Canada bushel. 
 
 WINE MEASURE. 
 
 1. In 3 hogsheads, how many gills ? Ans. 6048. 
 
 2. How many hogsheads in 6480 gills ? 
 
 Ans. 3 Hhds. 13 Gals. 2 Qts. 
 
 3. How many pints in 25 t'uns of wine ? Ans. 50400. 
 
 4. In 30876 gills, how many hogsheads 1 
 
 Ans. 15 Hhds. 19 Gals. 3 Qts. 1 Pt. 
 
 BEER MEASURE. 
 
 1. In 10 hogsheads 17 gallons, how many gills? 
 
 Ans. 17824. 
 
 2. How many firkins of ale, in 7624892 pints ? 
 
 Ans. 1 19188 A. Fir. 7 Gals. 2 Qts. 
 
 3. How many pints in 12 hogsheads, 15 gallons, 2 
 quarts ? Ans. 5308. 
 
 4. In 6420 quarts, how many firkins of Beer 1 
 
 Ans. 178 B. Fir. 3 Gals. 
 
 TIME. 
 
 1. How many minutes in 347 days 1 Ans. 499680. 
 
 2. In 57953 hours, how many weeks 1 
 
 Ans, 344 W. 6 D. 17 H. 
 
 3. How many seconds are there in 72 years, 10 days, 
 18 hours, 11 minutes, allowing 365 days and 6 hours to a 
 year ? " A ns. 2273076660. 
 
 4. How many days from the 20th of April to the 16th 
 of December following ? Ans. 240. 
 
 5. Suppose your age to be 16 years and 20 days, how 
 many seconds old are you, allowing 365 days and 6 hours 
 to the year 1 Ans. 506649600 Sec. 
 
 6. How many days from the birth of Christ, to Christ- 
 mas 1823, allowing the year to contain 365^ days ? 
 
 Ans. 665850J Days. 
 
 7. In a lunar month, how many seconds ? 
 
 Ans. 2551443 &. 
 
APPLICATION OF REDUCTION. 37 
 
 CIRCULAR MOTION. 
 
 1. In 6 signs of the zodiac, through which the sun 
 moves in half a year, how many seconds ? 
 
 Ans. 648000. 
 
 2. How many prime minutes in 360 degrees 1 
 
 Ans. 21600. 
 
 APPLICATION. 
 
 1. Four men brought each 70 sterling value in gold 
 into the mint ; how many guineas at 21s. each must they 
 receive in return 1 Ans. 266 guin. 14s. 
 
 2. A silversmith received 3 ingots of silver each weigh- 
 ing 54 ounces, with directions to make them into spoons 
 of 2 oz., cups of 5 oz., salts of 1 oz., and snuff boxes of 
 2 oz., and deliver an equal number of each ; what was 
 the number ? Ans. 16 of each, and 2 oz. over. 
 
 3. Suppose a ship's cargo from Bourdeaux to consist of 
 250 pipes, 130 hhds. and 150 quarter casks or 4 hhds. ; 
 how many gallons are there in all ; and, allowing every 
 pint to be a pound, what burden was the ship of? 
 
 Ans. 44415 gal., and the ship's burden ) 
 was 158 tons 12 cwt. 2 qrs. ) 
 
 4. In 15 pieces of cloth, each piece 20 yds, how many 
 French ells ? Ans. 200. 
 
 5. In 10 bales of cloth, each bale 12 pieces, and each 
 piece 25 Flemish ells, how many yards ? Ans. 2250. 
 
 6. The forward wheels of a wagon are 14^ feet in cir- 
 cumference, and the hind wheels 15 feet, 9 inches ; how 
 many more times will the forward wheels turn round than 
 the hind wheels, in running from Hallowell to Boston, it 
 being 174 miles ? Ans. 5028^ times. 
 
 7. How many times will a ship 97 feet, 6 inches long, 
 sail her length, in the distance of 12800 leagues, and 10 
 yards ? Ans. 2079508. 
 
 8. The sun's mean distance from the earth is 95,000,000 
 of miles ; and a cannon ball at its first discharge, flies 
 about a mile in 7^ seconds ; how long would a cannon 
 ball be, at that rate, in flying from the earth to the sun ? 
 
 Ans. 22yrs. 216days, 12h. 40min. 
 
 9. If a field be 36 rods long, and 24 rods wide, how 
 many acres does it contain ? Ans. 5 ac. 1 roo. 24 rods. 
 
 D 
 
68 FEDERAL MONEY. 
 
 10. How many strokes does a regular clock strike in 
 365 days or a year ? Ans. 56940. 
 
 11. How Jong will it take to count a million, at the rate 
 of 50 a minute Ans. 333h. 20m. or 13d. 2!h. 20m. 
 
 12. If the national debt of England amounts to 837 
 millions of pounds sterling ; how long would it take to 
 count this debt in dollars, (4s. 6d. sterling,) reckoning, 
 without intermission, twelve hours a day, at the rate of 
 50 dollars a minute ; and allowing 365 days to the year 1 
 
 Ans. 283 yrs. 38 days. 4 hours. 
 
 13. In 42 pigs of lead, each weighing 4cwt. 3qrs. how 
 many fother, at 19cwt. 2qr. ? Ans. 10 fother, 4^ cwt. 
 
 14. A gentleman has 20 hhds. of tobacco, each 8 cwt. 
 3 qrs. 14ft), and wishes to put it into boxes containing 
 70 Ib. each ; I demand the number of boxes he must get, 
 at 25ft to the qr. ? Ans. 254. 
 
 15. How many coats can be made out of 36f yds. of 
 broadcloth, allowing If yds. for a coat 1 Ans. 21. 
 
 16. A man would ship 720 bushels of corn, in barrels 
 which will hold 3 bus. 3 pks. each ; how many barrels 
 must he get ? Ans. 192. 
 
 1 7. How many pints, quarts, and two quarts, of each an 
 equal number, may be filled from a pipe of wine ? 
 
 * . Ans. 144. 
 
 18. Three fields contain, the first 7 acres, the second 
 10 acres, and the third 12 acres, 1 rood; how many shares 
 can they be divided into, each share to contain 76 perches? 
 Ans. 61 shares, and 44 perches over. 
 
 FEDERAL MONEY,* 
 
 THE denominations of Federal Money, like figures in 
 whole numbers, increase in a tenfold proportion, begki- 
 ing with mills, of which 
 
 10 make - 1 Cent, marked m. c. respectively. 
 
 10 Cents - 1 Dime, - - - d. 
 
 10 Dimes - 1 Dollar, - - Doll or $. 
 
 10 Dollars I Eagle, - - - E. 
 
 * Federal Money ought, in strict propriety, to be treated of 
 after decimal fractions ; but usefulness, (as fractions are not al- 
 
FEDERAL MONEY. 39 
 
 In the money of account the dollar is considered as the 
 unit : all other denominations being valued according to 
 their distance from the dollar's place. A point or comma 
 must be placed after the dollars to separate them from the 
 lower denominations ; then the first figure at the right of 
 the comma is dimes, the second cents and the third 
 mills ; but in reckoning, the two first are called so many- 
 cents, using the dimes for the tens' place of cents.* 
 When the cents in any sum are less than 10, a cipher 
 must be put in the place of dimes, or tens' place of the 
 cents, before any operation is performed, 
 
 ADDITION OF FEDERAL MONEY. 
 
 RULE. Place the numbers according to their value, 
 dollars under dollars, cents under cents, <fcc. and add as 
 in whole numbers, placing the comma in the sum directly 
 under the commas above. 
 
 ways understood) requires, and its simplicity and near alliance to { !\f[ 
 whole numbers, will admit it in this place. 
 
 The coins of the United States are three of gold, six of silver 
 and two of copper. The gold coins are called an eagle, half 
 eagle and quarter eagle ; the silver, a dollar,half dollar, quarter 
 dollar, double dime, dime and half dime ; and the copper, a cent 
 and half cent. 
 
 The weight of the eagle is 11 penny weights and 6 grains ; the 
 weight of the dollar 17 pennyweights 8 grains; of the dime, 1 
 pennyweight 17 3-5 grains ; of the cent 8 pennyweights 16 grains. 
 The standard for gold coin is eleven parts fine gold and one part 
 alloy; the alloy consisting of silver and copper. The standard 
 for. silver is 1435 to 179 alloy ; the alloy being wholly of copper. 
 Alloy is used in gold or silver to harden it. 
 
 When either gold or silver is finer or coarser than the standard, 
 the variation from the standard is estimated by carats and grains 
 of a carat in gold, and by pennyweights in silver. 
 
 NOTE. A carat is not any certain quantity or weight^ but^- of 
 any weight or quantity ; which minters and goldsmiths divide 
 into four equal parts called grCtins of a carat. 
 
 * Any sum of this money may be read differently ; either wholly 
 in the lowest denomination, or partly in the higher and partly in 
 the lowest. Thus the sum $24, 367 may be read 24367 mills ; or 
 2436 cents. 7 mills ; or 243 dimes, 6 cents and 7 mills ; or 24 dol- 
 lars, 3 dimes, 6 cents and 7 mills ; or 2 eagles, 4 dollars, 3 dimes, 
 6 cents and 7 mills ; or 24 dollars, 36 cents and 7 mills ; but the 
 last is the usual method. 
 
FEDERAL MONEY, 
 
 EXAMPLES. 
 
 I. 
 
 2. 
 
 3. 
 
 c. m. 
 4612,39 4 
 5746,29 6 
 6154,34 7 
 2410,61 8 
 3917,46 5 
 
 $ c. m. 
 274,62 1 
 125,48 6 
 612,59 4 
 431, 51 3 
 246,34 9 
 
 $ c. m. 
 489,20 6 
 217,16 5 
 6,74 9 
 12,81 6 
 7,10 
 
 Sum. 22841,120 
 
 18228,72 6 
 
 Proof. 22841,12 
 
 
 
 4. 
 
 5. 
 
 6. 
 
 <s> 
 $>. c. 
 
 3792,47^ 
 684,16 
 59,76 
 1246,23} 
 
 $. c. m. 
 76,28 1 
 39,46 2 
 57,19 7 
 68,49 8 
 
 9. c. 
 180,20 
 29, 
 6,17 
 31,42 
 
 
 APPLICATION. 
 
 l Suppose that B owes A $75, 17c ; C owes 15c. 4m ; 
 D owes $21, 13c. 6m ; E owes 9c. ; F owes $796, 3c. ; 
 and G owes $17,13c. ; what is due to A from all of 
 them : Ans. $909,71c. 
 
 2. There is a gallant ship just returned from the Indies, 
 which is herself worth $12l45,86c. ; and one quarter of 
 her cargo is valued at $25411,65c. ; pray tell me the 
 value of the ship and cargo. Ans. $113792,46c. 
 
 SUBTRACTION OF FEDERAL MONEY. 
 
 RULE. Place the less sum under the greater in the 
 same manner as in addition. Subtract as in whole num- 
 bers, and place the comma directly under those above. 
 
FEDERAL MONEY. 
 
 41 
 
 1. 
 
 $ c. m. 
 From 4612,60 5 
 Take 2904,71 6 
 
 EXAMPLES. 
 2. 
 
 $ c. 
 
 41,20 
 
 3,97 
 
 3. 
 
 ni. 
 317,61 1 
 149,72 1 
 
 Rem. 1707,889 
 
 Proof 4612,60 5 
 
 4. 
 
 $ c, m. 
 
 2910,70 2 
 
 827,06 4 
 
 APPLICATION. 
 
 1. Suppose that my rent for three months is $268, and 
 that I have paid for taxes $58,16c., and for several repairs 
 $73,85^c. ; what have I to pay of my quarter's rent 1 
 
 Ans. $135,98*c. 
 
 2. Jack Hatchway received prize money to the amount 
 of $1000 ; he then laid out $411,410. for a span of fine 
 horses ; $123,40c. for a suit of new clothes, and a gold 
 watch ; and $359, 50c. were lost in lottery gambling ; 
 what will he have left, after he has paid his landlord's bill, 
 which is $S5,llc. ? Ans. $20,5Sc. 
 
 MULTIPLICATION OF FEDERAL MONEY. 
 
 . RULE Multiply as in whole numbers, and place the 
 comma as many figures from the right hand in the pro- 
 duct as it is in the multiplicand. 
 
 EXAMPLES. 
 
 1. 2. 
 
 $ci> 
 c. $ c. m* 
 
 Multiplicand, 4769,67 4276,96 7 
 
 Multiplier, 36 48 
 
 Product. 
 
 2861802 
 1430901 
 
 171708,12 
 D2 
 
 205294,41 
 
42 FEDERAL MONEY. 
 
 $ c. m. $ c. m. 
 
 3. Multiply 67,48 2 by 5 Product 337,41 
 
 4. 76,43 4 305,72 
 
 5. 3,164 9 28,476 
 
 6. ,78 1 12 9,37 2 
 
 7. 1,06 45 47,70 
 
 8. 3,16 150 474,00 
 
 9. 4,25 598 2541,50 
 
 10. 4,96 3 347 1722,16 1 
 NOTE. To multiply by , *, -| , &c. Take -J, \. |, &c. 
 
 of the multiplicand first, and set it down beneath the line 
 as a product; then multiply by the whole number, set- 
 ting the product or products below that of the fraction, 
 and add all together. 
 
 11. 10,50 14^ 152,25 
 
 12. 1,20 84J 101,10 
 
 13. 2,40 26f 64,20 
 
 APPLICATION. 
 
 1. What will 120 yards of damask come to, at $12, 
 5c. per yard ? Ans. $1446. 
 
 2. Find the amount of the following bill of Parcels. 
 
 Hallowell, Nov. 1, 1829. 
 Mr. Peter Paywell, 
 
 Bought of Francis Fairdealer ; 
 
 28 fb> of Green Tea, at $2, 15c. per ft. $ c. 
 41' ft. of Coffee, 0, 15 
 
 34 ft. of Loaf Sugar, 0, 19 
 
 13 cwt. of Malaga Raisins, 7, 31 per cwt. 
 
 35 firkins of Butter, 7, 14 per fir. 
 27 pairs of Worsted Hose, 1, 4 per pair, 
 94 bushels of Oats, 0, 33 per bush. 
 
 29 pairs of Men's Shoes, 1, 12 per pair, 
 
 Amount, $509,32c. 
 Received payment in full, 
 
 Francis Fairdealer. 
 
 A SHORT RULE, 
 
 To know, mentally, the value of 100 pounds of any article 
 in Federal Money, when the price of lib. is given. 
 
 RULE. Cnil the cents in the price of 1 pound, dollars, 
 and that sum v/Ul be the value of 100 pounds of the arti- 
 
FEDERAL MONEY. 43 
 
 cle. If there be several hundred pounds of it, multiply 
 the value of 100ft. thus found, by the number of hun- 
 dreds and you will have the answer accordingly. The 
 whole may be done by a single glance of the mind. 
 Parts of a cent in the price of 1 pound, will be the same 
 parts of a dollar in the price of 100 pounds. 100ft. at 
 1 cent per ft.=100 cents = l dollar. 
 
 Therefore, lOOjfr. of beef, at 4 cents a ft. will come 
 to 400 cts. =$4. 
 
 What will 500ft. of pork come to at 8 cts. a ft. ? 
 
 Your mind tells you 8 cts. a ft. is $8 a 100 pounds and 
 8 multiplied by 5 is 40 ; of course the answer is $40. 
 
 DIVISION OF FEDERAL MONEY. 
 
 RULE. Write the number and divide as in simple di- 
 vision. The quotient will be of the same denomination 
 as the lowest of the dividend. Or, when you have divid- 
 ed the dollars of the dividend, put the comma or point 
 in the quotient ; if the dollars be less than the divisor, 
 put the point down at first. 
 
 EXAMPLES. 
 1. 2. 3. 
 
 $. c. m. $. c. $. c. m. 
 
 6)47,26 2 8)6914,24 1 1 )7,49 1 
 
 17,87 7 
 
 4. 
 
 - f . c. m. $. c. m. 
 
 237)6742,27 1(28448+ mills, or 28,44 8 and 95 Rem. 
 
 5. 6. 7. 
 
 $. c. w. $. c. m. $. c. m. 
 
 387)753,35 7( 359)259,23 7( 475)74,10 0( 
 
 APPLICATION. 
 
 1. If 131 yards of Irish linen cost $49,78c., what is it 
 per yard 1 Ans. 38c. 
 
 2. If 140 reams of paper cost $329, what is that per 
 ream ? Ans. $2,35c. 
 
 3. If a reckoning of $25,41c. be paid in equal shares 
 by 14 persons, what do they pay apiece ? Ans. $l,81^c. 
 
 4. If a man's wages be $235,80c. a year, what is that 
 a calendar month ] Ans. $19,65c. 
 
44 FEDERAL MONEY. 
 
 5. The salary of the President of the United States, is 
 twenty-five thousand dollars a year ; what is that a day 1 
 
 Ans. $68,49c.+ 
 
 6. If the amount of the Public Debt of the United 
 States, be $91,680,090 ; how much would that be for each 
 person to pay, allowing the number of inhabitants in the 
 United States to be eleven millions? Ans. 88,33c. 4m. 4- 
 
 7. Divide $57 into 120 equal shares. Ans. 47Jcents. 
 
 A SHORT RULE. 
 
 When the value of lOOjfo of any article is given, to 
 find, mentally, what it is per ffo. 
 
 RULE, Call the dollars in the price of lOOjfj. cents, 
 and that is the answer. If the value of several 100}^. is 
 given, first divide, in your mind, the dollars in the price 
 of the hundreds, by the number of hundreds, and the 
 quotient will be the price of one hundred, in dollars, 
 which call cents, as before directed, and that will be the 
 answer. When the price of the 100ft,. in either case, 
 contains parts of a dollar, the parts of a dollar become 
 parts of a cent in the price of the ]fo. 
 
 100ft) valued at 85=100^5 valued at 500 cents ; and 
 500 cents-r- 100=5 cents. Therefore if lOOffc. of beef 
 cost 6 dollars, or 600 cents, it cost 6 cts. a ffe. 
 
 What will a ffo of pork be worth, when SOOJfo are 
 worth $48. 
 
 Your mind tells you 800fc for $48 will be lOOffe for 
 $6, or $48 divided by 8, the number of hundreds, will 
 give $6 for the quotient, or price of 1 hundred^; of course 
 the price of a pound, or the answer, is 6 cents. 
 
 REDUCTION OF FEDERAL MONEY. 
 
 To reduce Dollars to Cents and Mills. 
 Multiply the dollars by 100 for cents, and the cents by 
 10 for mills or to the dollars annex two ciphers for cents 
 and three for mills. 
 
 To reduce Mitts and Cents to Dollars. 
 Divide the mills by 10, and the quotient will be cents ; 
 divide the cents by 100, and the quotient will be dollars ; 
 
FEDERAL MONEY. 45 
 
 or if the number be cents, point off two ; and if mills, 
 three figures on the right hand; then the figures on the 
 left hand of the comma will be dollars, the two first on 
 the right hand will be cents, and the third, if any, will 
 be mills. 
 
 EXAMPLES. 
 
 1; Reduce 674 dollars to cents and mills. 
 674 
 100 
 
 67400 Cents 
 10 
 
 674000 Mills. 
 
 Or 67400 Cents. ) A 
 674000 Mills. J * 
 
 2. How many dollars in 642179 mills ? 
 10)642179 
 
 100)64217-9 
 
 $642-17-9 Or $642 17c. 9m. Ans. 
 
 3. How many mills in 47692 dollars ? Ans. 47692000. 
 
 4. In 46791 cents, how many dollars ? Ans. $467,91c. 
 
 5. In 6421796 mills, how many dollars, cents and mills ? 
 
 Ans. $6421 79c. 6m. 
 
 To reduce New-England currency to federal Money. 
 CASE 1. If the sum consist of pounds only, annex 
 three ciphers to it and divide by 3 ; the quotient will be 
 the answer in cents.* 
 
 EXAMPLES. 
 
 1. Reduce 3762 to Federal Money. 
 3)3762000 
 
 1254000 Cents, or $12540 Ans. 
 
 2. Reduce ,471 to Federal Money. Ans. $1570. 
 
 3. Reduce .37 to dollars and cents. Ans. $123 33c. 
 
 * As a dollar is 2 iy orf^- of a pound, it is plain that annexing 
 a cipher to the pounds, and dividing- by 3, will give a quotient in 
 dollars ; and annexing- other ciphers, and dividing- by 3 will give 
 tenths, hundredths, &c. of a dollar ; or dimes, cents, &c. 
 
46 FEDERAL MONEY. 
 
 CASE II. If pounds and shillings are given, to the 
 pounds annex half the number of shillings, and two ci- 
 phers, if the number of shillings be even ; but if the 
 number be odd, annex half the even number, and then 5 
 for the odd shilling, and one cipher, and divide by 3 ; the 
 quotient is the answer in cents. 
 
 EXAMPLES. 
 
 1. Reduce 64 16s. to dollars and cents. 
 
 3)64800 
 
 Ans. 21600cs. or $216. 
 
 2. How many dollars in ,41 14s. ? Ans. $139. 
 
 3. In <26 Is. how many dollars and cents ? 
 
 Ans. g86,83c*s. 
 
 4. How many dollars in 1 17s.? Ans. $6,16fc*s. 
 
 CASE III. If there are shillings, pence, &c. in the 
 given sum, annex for the shillings as before, and to these 
 add the farthings contained in the pence and farthings ; 
 observing to increase their number by 1 when they exceed 
 12, and by 2 when they exceed 36 ; and divide as before. 
 
 EXAMPLES. 
 
 1. Reduce c34 165. 4Jrf. to Federal Money. 
 
 3)34819 
 
 Ans. 11606^5. or $116,06^5. 
 
 2. In c200l Is. 3d. how many dollars ? 
 
 Ans. $6670,21fc*s. 
 
 3. In c591 11s. 9%d. how many dollars ? 
 
 Ans. $1971,96fc*s. 
 
 To reduce FEDERAL MONEY to New-England currency. 
 
 CASE I. When the sum is dollars only, multiply by 3 : 
 and double the product of the first figure for shillings, 
 and the rest of the product will be pounds. 
 
 EXAMPLES. 
 
 1. Reduce 473 dollars to New-England currency. 
 473 
 3 
 
 Ans. c141 18s. 
 
FEDERAL MONEY. 47 
 
 2. How many pounds, &c. in 579 dollars? 
 
 Ans. ,173 14s. 
 
 CASE II. When there are cents in the given sum, 
 multiply the whole by 3, and cut off three figures of the 
 product to the right hand as a remainder ; multiply the 
 remainder by 20, and cut off as before : proceed in the 
 same manner through the several parts of a pound, and 
 the numbers standing on the left hand make the answer 
 in the several denominations. 
 
 NOTE. If there be mills, cut off four figures, and 
 proceed as before. 
 
 EXAMPLES. 
 
 1. Reduce $376,27cs. to New-England currency. 
 376,27 
 3 
 
 .112,881 
 20 
 
 s. 17,620 
 12 
 
 d. 7,440 
 4 
 
 qr. 1,760 
 Ans. <112 17s. 
 
 2. Reduce $609,88cfc>. to New-England currency. 
 
 Ans. <182 19s. 
 
 3. How many pounds, shillings, &c. in $429,21cfc. 5 
 mills 1 
 
 429,215 
 3 
 
 ,128,7645, <fcc. Ans. ,128 15s. 
 
48 COMPOUND ADDITION. 
 
 PRACTICAL QUESTIONS IN FEDERAL MONEY. 
 
 1. Having borrowed one hundred dollars, arid paid at 
 one time seventy dollars, and at another time sixteen dol- 
 lars, seven cents ; what is still due ? Ans. $13,93c?s. 
 
 2. What will 25 bushels of corn come to, at 92 cents 
 per bushel 1 Ans. $23. 
 
 3. What will 376 pounds of butter come to, at 18 
 cents per pound ? Ans. $67,68cs. 
 
 4. What will 39 bushels of wheat come to, at 1 dollar 
 75 cents per bushel ? Ans. $68,25cs. 
 
 5. If Iron cost 6 dollars 50 cents per cwt., what is it 
 per pound, 25jfc. to the qr. ? Ans. 6cts. 
 
 6. If 240 pounds of pork come to 28 dollars 80 cents, 
 what is it per pound ? Ans. 12c;te. 
 
 7. Borrowed 607 dollars 20 cents ; paid ,127 16s. 9d. ; 
 what is the balance ? Ans. $181,7cs. 
 
 8. Lent 400 dollars ; received 150 bushels of wheat at 
 10s. per bushel, and 200 pounds of butter at 17^ cents 
 per pound, in payment : how much is still due ? 
 
 Ans. $115. 
 
 9. What will 55 yards of linen come to, in Federal 
 Money, at 3s. 9d. per yard, New-England currency 1 
 
 Ans. $34,37rts. 5mitts. 
 
 10. What will 36 yards of broadcloth come to, in New- 
 England currency, at 6 dollars 25 cents per yard ? 
 
 Ans. <67 105. 
 
 11. What will 7 tons of hay come to, at 16 dollars 75 
 cents per ton ? Ans. $117 25rts. 
 
 12. What is the price of one cwt. of hay at 16 dollars 
 75 cents per ton ? Ans. 83cts. 
 
 COMPOUND ADDITION. 
 
 COMPOUND ADDITION in Arithmetic, teaches to collect 
 several numbers of different denominations into one sum ; 
 as, pounds, shillings and pence, &c. ; tons, hundreds, 
 and quarters, &c. 
 
 RULE. Place the numbers, so that those of the same 
 denomination may stand directly under each other, ac- 
 
COMPOUND ADDITION. 49 
 
 cording to what you were told in Simple Addition. Add 
 the first column or lowest denomination together, as in 
 Simple Addition, also. Find how many units of the next 
 higher denomination are contained in the sum, by divid- 
 ing it by so many of this name as make one in the next 
 greater. Set down the remainder, or overplus, under the 
 column added, and carry the ones or units (the quotient) 
 to the next denomination, whose sum you must find, and 
 proceed with as before ; and so on through all the denom- 
 inations to the highest, which add as in Simple Addition 
 setting down its whole sum.* 
 
 EXAMPLES. 
 
 MONEY. 
 
 . s. d. . s. d. qr. . s, d. 
 
 4210 1611 97 12 6 2 63 9 4J 
 
 1729 17 9 19 14 5 3 8112 7 
 
 56 11 34- 46 17 9 1 27 16 
 
 207 18 6f 221910 2 1614 
 
 14 6 57 1 2 41 9 5 
 
 Sum. 6219 10 6f 
 2008 13 7! 
 Proof621910 6 
 
 APPLICATION. 
 
 1. Bought a quantity of goods for c125 10s. ; paid for 
 truckage forty-five shillings, for freight seventy-nine shil- 
 lings and sixpence, for duties thirty-five shillings and ten 
 pence ; and my expenses were fifty-three shillings and nine 
 pence ; what did the goods stand me in 1 Ans. <136 4s. \d. 
 
 * The reason of this rule is evident ; for, in addition of money, 
 as 1 in the pence, is equal to 4 in the farthings 1 in the shil- 
 lings to 12 in the pence and 1 in the pounds, to 20 in the shil- 
 ling's, to carry as directed is, therefore, only to arrange properly 
 the money, arising from each column, in the scale of denomina- 
 tions : and this reasoning- will hold just in the addition of com- 
 pound numbers of any kind whatever. 
 E 
 
50 COMPOUND ADDITION. 
 
 2. A prize being sold, and the sum divided equally 
 among the captors, who were 6 in number, each man re- 
 ceived two hundred and forty pounds, thirteen shillings 
 and seven pence ; what did the prize cost the purchaser ? 
 
 Ans. ,1444 Is. Qd. 
 
 TROY WEIGHT. 
 
 jfe. oz. dwt. gr. ft. oz. dwt. gr. 
 
 671 10 16 13 47 6 11 17 
 
 392 8 9 21 56 11 15 19 
 
 249 7 12 12 63 9 8 22 
 
 516 4 3 7 26 5 19 16 
 
 627 5 17 16 21 6 12 14 
 
 APPLICATION. 
 
 A gentleman bought of a silversmith, dishes to the 
 weight of 23ft 6oz. 5dwt. ; plates 41jfo. 7oz. 17dwt. ; 
 spoons 12|fc. 15dwt, ; salts 2ft. 7oz. ; waiters 13ft. ; and 
 tankards 7ft. 17dwt. ; what weight of plate did he buy 
 in all 1 Ans. 99ft. lOoz. 14dwt. 
 
 AVOIRDUPOIS WEIGHT. 
 
 Ton. cwt. <?r.ft. oz. dr. Ton. cwt. qr. ft 
 
 371 12 1 20 10 13 42 19 3 24 
 
 123 14 2 15 6 7 89 10 2 13 
 
 407 9 3 12 11 9 16 8 1 17 
 
 513 13 26 9 15 27 14 22 
 
 624 15 1 17 6 11 50 3 2 15 
 
 Here 25ft. a qr. 
 
 APPLICATION. 
 
 1. A merchant buys four bags of hops, of which No. 1, 
 weighs 2cwt. 2qr. 10ft. No. 2, 2cwt. Iqr. 16ft. No. 3, 
 2cwt. 24ft. No. 4. Iqr. 16ft. He buys also a couple of 
 pockets of hops, which weigh 58Jft. each. How many 
 hundred weight has he to pay carnage for, on bringing 
 them to town ? Ans. 8cwt. 2qr. 15ft. 
 
 2. A country shopkeeper buys of a merchant in Hallo- 
 well, teas weighing 3qrs. 14ft. ; coffee, Iqr. 23ft. ; sugars, 
 
COMPOUND ADDITION. 51 
 
 3ewt. 2qr. 5ft. ; spices 2qr. 3ft. 13oz. ; hops 13cwt. Iqr. 
 24ft. ; and several other things to the weight of 3cwt. 17ft 
 7oz. ; for what Weight has he to pay carnage on bringing 
 them home, 25ft. a qr. ? Ans. 22cwt. 12ft. 4oz. 
 
 APOTHECARIES' WEIGHT, 
 
 ft- 3- 9-^r. ft- I- 3- 9-s*-. 
 
 26 10 7 2 13 17 9 4 1 14 
 
 54 7 2 1 12 55 10 6 2 10 
 
 76 8 3 6 14 61 4 2 1 9 
 
 83 9 4 2 6 92 11 5 18 
 
 41 6 1 19 21 6 3 1 17 
 
 APPLICATION. 
 
 An apothecary made a composition of 5 ingredients, 
 the 1st of which weighed 13ft. 7oz. ; the 2d, lloz. 7dr. 
 ISgr. ; the 3d, 7ft. 2scr. ; the 4th, lift. 3dr. Iscr. ; and 
 the 5th weighed 15ft. 5oz. 7gr. ; what was the weight 
 of the whole 1 Ans. 48ft. 3dr. Iscr. 
 
 CLOTH MEASURE. 
 
 Yd. qr. na. E.E. qr. na. E.Fr. qr. na. E.FL qr.na. 
 
 46 1 2 ' 74 2 3 86 5 2 29 1 2 
 
 12 3 3 51 4 1 61 4 1 17 2 3 
 
 62 1 1 24 1 2 52 3 9 20 1 2 
 
 83 2 56 3 1 24 2 1 75 1 
 
 41 1 31 1 2 10 3 46 1 3 
 
 APPLICATION. 
 
 1. Having bought four parcels of cloth, the first of 
 which contains 25yds. 3qrs. ; the 2d, 37yds. 2qrs. 3na. ; 
 the 3d, 14yds. Ina. ; and the 4th, 23yds. ; I desire to be 
 informed by you how many yards are in them all ? 
 
 Ans. 100yds. 2qrs. 
 
 2. I have six parcels of cloth ; the first contains 3 E. 
 Fl. and 3na. ; the 2d, 14yds, Iqr. ; the 3d, 15E. E. 2qr. 
 2na. the 4th, Iqr. 3na. ; the 5th, 19E. Fr. ; and the 6th, 
 255yds. ; how many yards are there in all 1 
 
 Ans. 320yds. 
 
52 COMPOUND ADDITION. 
 
 LONG MEASURE. 
 
 Deg. mi.fur.pol ft. in. bar. Mi. fur. rod. yd. ft. 
 
 2i 17 2 26 12 10 2 62 1 36 4 2 
 
 46 58 6 19 14 6 1 75 3 24 3 1 
 
 18 62 4 31 10 7 49 5 12 2 2 
 
 21 19 7 14 8 8 2 14 4 17 1 
 
 62 37 1 29 7 *6 1 25 2 10 3 2 
 
 APPLICATION. 
 
 1. From A to B is 3mi. 2fur. 7pol. ; from B to C is 17 
 mi. 13pol. ; from C to D is 7fur. and from D to E is 5 
 mi. 33pol. ; what is the distance between A and E 1 
 
 Ans. 26mi. 2fur. 13pol. 
 
 2. A person rode four days ; on the 1st day he went 39 
 mi. 6fur. ; the 2d day, 46mi. 24pol. ; the 3d, 60mi. 4fur. 
 39pol. ; and the 4th day he went but 37mi. 6fur. ; what 
 was the whole distance of his journey ? 
 
 Ans. I84mi. Ifur. 23pol. 
 
 LAND OR SQUARE MEASURE. 
 
 Acre. rood. rod. yd. ft. in. Acrc.rood. rod. 
 
 271 1 27 16 4 110 21 2 37 
 
 424 2 3i 21 2 96 52 1 24 
 
 512 3 16 10 7 71 28 1 16 
 
 327 2 21 9 1 120 43 3 31 
 
 146 1 12 21 6 74 65 20 
 
 APPLICATION. 
 
 1. There are 5 lots of land, the 1st of which measures 
 I3ac. 3roo. I4rods; the 2d, 27ac. 29rods ; the 3d, I9ac. 
 Iroo. ; the 4th, 3roo. 34rods ; and the 5th, 45ac. 2roods, 
 1 1 rods ; what ground do they all contain ? 
 
 Ans. I06ac. 3roods, 8rods. 
 
 2. A gentleman dividing his estate among 4 sons, gave 
 the 1st, I50acres, 1 rood, 39rods ; the 2d, 100acres,25rods; 
 the 3d, 75acres : and the 4th, 55acres, Irood, I6rods; 
 how large was the whole farm ? Ans. 381acres. 
 
COMPOUND ADDITION. 53 
 CUBIC OR SOLID MEASURE. 
 
 T.round. ft. inch. T.liewn. ft. Cord. ft. 
 
 21 36 476 67 30 36 102 
 
 62 17 965 24 27 6i 90 
 
 47 19 8.16 62 14 52 76 
 
 31 28 1146 39 36 9 120 
 
 19 34 1452 17 20 12 34 
 
 AP PLICA TION. 
 
 1. Required the solid contents of four pieces of round 
 timber, the first of which measured 2tons, 39ft. 94in. ; the 
 2d, 4tons, 1695in. ; the 3d, 35ft. 1183in. ; and the 4th, 
 Iton, 25ft. Ans. 9T. 20ft. 1244in. 
 
 2. Bought of A 14cor. 1727sol. in. of wood ; of B 
 19cor. 127sol. ft, ; of C 7eor. 98ft. llOlin. ; of D 9cor. ; 
 and of E 63ft. 1210in. ; how much did I buy in all ? 
 
 Ans. 51c. 34ft. 5S2in. or 51c. 2|ft.+Wood Measure, 
 
 DRY MEASURE. 
 
 Chal.bus.plf. qt.pt. Bus.pk. gal. qt.pt, 
 
 39 4 1 6 I 62 2 1 3 1 
 
 57 6 3 4 79 3 2 1 
 
 61 7231 42 2 1 1 1 
 
 52 4 2 5 1 17 1 1 2 
 
 42 1 1 2 1 86 1 1 
 
 APPLICATION. 
 
 1. A corn-merchant sends over the sea, of wheat 130 
 bus. 3pks. Igal. Ipt. ; of oats 290bus. ]gal. 3qts. ; of rye 
 he has sent 300bus. 3qts. ; of pease 80bus. 3pks. ; and of 
 beans 50bus. ; for what number of bushels does he pay 
 freight 1 Ans. 851 bus. 3pk. Igal. 2qt. Ipt. 
 
 2. How many chaldrons of coals are there in four 
 Joads, the 1st containing l4chal.27bus.2pk. Iqt. ; the 2d, 
 9chal. 3pk. ; the 3d, 29chal. 3qts. ; arid the 4th, 35chal. 
 7bus. 2pk. 4qts. ? Ans. 88chal. 
 
54 COMPOUND ADDITION. 
 
 WJNE MEASURE. 
 
 Tun.hhd.gal qt. pt.gilL Jfhd. gal. qt. 
 
 2 . 1 42 1 3 68 31 3 
 
 5 3 51 2 1 1 47 59 1 
 7 2 60 1 1 2 29 48 1 
 9 1. 26 1 3 68 37 
 
 6 1 17 3 1 1 47 18 1 
 
 APPLICATION. 
 
 1. A gentleman bought oC a wine-merchant, of port 
 wine, 1 tun, 3hhds. ; of claret, 3hhds. 47gal. ; of moun- 
 tain, Ihhd. 5gal. ; and of Lisbon, 2hhds. 23gal. : what 
 quantity did he buy in all 1 Ans. 3tuns, 2hhds. I2gal. 
 
 2. Imported from Cadiz, Lisbon, Oporto, and Madeira, 
 4 lots of wine, the 1st of which contained 3tuns, 3hhds, 
 61gal. 2qt. Ipt. ; the 2d, 2pipes, Ihhd. 48gal. 3gills ; 
 the 3d, 2hhds. 41 gal. 1 gill; and the 4th, 6pipes, 3hhds. 
 38gal. Iqt. : how much did I have in all ? Ans. 10 tuns. 
 
 ALE OR BEER MEASURE. 
 
 Hhd.gaLqt. A.fr.galqt. Butt.bblgalqt. 
 
 416 29 3 69 7 2 25 1 24 1 
 
 34 17 2 43 5 1 36 2 30 2 
 
 178 15 1 53 4 I 81 16 
 
 315 47 1 46 3 3 74 1 25 1 
 
 351 34 1 29 1 1 98 1 18 1 
 
 APPLICATION. 
 
 1. A beer-brewer has sent into the country, ale, as fol- 
 lows, viz. at one time 3hhds. I4gal. 3qt. : at another, 2 
 hhds. I6gal. ; at another, I4hhds. 27gals. 2qts/; and at 
 another, 6hhds. 3qts. : how much was sent in all ? 
 
 Ans. 26hhds. 5gal. 
 
 2. Bought 4 lots of beer;,*he 1st contained 3 butts, I 
 bbL3qts. ; the 2d, Ibutt, 34gaJ. Iqt.; the 3d, 2bbls. 2 
 qts. ; and the 4th, 5butts, Ibbl. 2qts%; what did I buy in 
 all? Ans. lObutts, 2bbls. 
 
COMPOUND ADDITION. 55 
 
 Fr. 
 37 
 
 46 
 
 58 
 61 
 94 
 
 m. 
 10 
 9 
 7 
 5 
 12 
 
 w. 
 2 
 3 
 1 
 
 3 
 
 a. 
 
 6 
 2 
 4 
 5 
 1 
 
 A. TW. 
 1731 
 1447 
 13 19 
 1250 
 21 39 
 
 TIME. 
 s. 
 37 
 25 
 31 
 47 
 56 
 
 Yr. days, 
 41 276 
 81 310 
 47 163 
 2t 360 
 72 176 
 
 h. 
 20 
 17 
 8 
 19 
 14 
 
 m. 
 30 
 48 
 29 
 50 
 33 
 
 APPLICATION. 
 
 1 . Peter's father was 28 years old, (reckoning 1 3 months 
 to 1 year, and 28 days to 1 month,) when Peter was born ; 
 betwixt Peter and John were 2yrs. 10m. I6d. ; between 
 John and James, one year and eleven months ; and between 
 James and Job, 3yrs. 7m. 25d. ; when Job is I6yrs. 9m. 
 27d. old, how old is their father 1 Ans. 53yrs. I2days. 
 
 2. Four memorable events occurred, between the 1st 
 and 2d of which were 2lyrs. 236d. 7h. 40m. 50s. ; be- 
 twixt the 2d and 3d, 48yrs. l9h. 53s. ; and between the 3d 
 and last, 30yrs. 128d. 3h. 18m. 17s. ; now, allowing 365d. 
 and 6 hours to the year, I wish to know what 'period of 
 time elapsed between the first and last of those events 1 
 
 Ans. 100 years. 
 CIRCULAR MOTION. 
 
 g O t ' H 9 f If 
 
 2 15 42 57 41 54 39 
 5 19 55 33 64 27 21 
 1 22 47 28 52 36 42 
 
 3 10 20 12 78 19 16 
 1 9 11 24 93 25 34 
 
 APPLICATION. 
 
 1. If the sun's motion in the zodiac be one day, 1 1' 
 10"; on another, the same; on the next, 1 1' 11"; on 
 the next, 1 1' 10" ; and on the next two, 1 1' 11" each 
 day ; what distance does he move in the six days 1 
 
 Ans. 6 7' 3". 
 
 2. If the moon's motion in the signs be, on one day, 13 
 28' 4"; on another, 13 2' 23"; on another, 12 40' 
 1"; on another, 12 2i> 47": on another, 12 ?' 50" ; 
 and on another, 11 56' 7"; what distance does she 
 move in the six days 1 < Ans. 2& 15 38' 12*, 
 
66 COMPOUND SUBTRACTION. 
 
 COMPOUND SUBTRACTION. 
 
 COMPOUND SUBTRACTION is finding the difference be- 
 tween two numbers, of which one or both are compound. 
 
 RULE. Set the less number under the greater, as di- 
 rected in Compound Addition. Then, beginning at the 
 least denomination, subtract the under number of each 
 from the upper, writing their respective remainders below 
 them. But if the under number of any of the denomi- 
 nations be greater than the upper, add so many to the 
 upper as make one of the next higher denomination ; then 
 take the under number from that sum, writing down the 
 remainder as before, and carry or add one to the under 
 number of the next higher denomination before you sub- 
 tract it. The method of proof is the same as in Simple 
 Subtraction. 
 
 EXAMPLES. 
 
 MONEY. 
 
 . s. d. . s. d. qr. . 5. d. 
 
 From 691 12 6^ 34 11 4 1 81 17 6 
 
 Take 284 15 9f 17 14 10 3 21 12 4 
 
 Rem. 406 16 82 
 
 Proof. 691 12 6 
 
 AP PLICA TION. 
 
 1. What sum added to .17 11s. 8jd. will make ,100 ? 
 
 Ans. ,82 8s. 3d. 3qrs. 
 
 2. Borrowed .50 10s. ; paid again, at one time, ,17 
 11s. 4d. ; at another, <9 4s. 8d. ; at another, 7 9s. 6d. ; 
 and at another, 19s. 6^d. ; how much remains unpaid ? 
 
 Ans. 15 4s. 
 
 TROY WEIGHT. 
 
 jfc. oz. dwt. gr. ft. oz. dwt. gr. 
 
 39 8 14 16 71 9 16 11 
 
 16 10 10 19 35 1 17 20 
 
 APPLICATION. 
 
 Sent to a Silversmith 9ffe. 8oz. 14gr. of silver to be 
 wrought ; he makes me four dozen of spoons, weighing 
 8oz. 6dwt. 21gr. ; how much silver is left 1 
 
 Ans. lloz. 13dwt 17gr. 
 
COMPOUND SUBTRACTION. 57 
 
 AVOIRDUPOIS WEIGHT. 
 Ton. cwt. qr. lb. Cwt. qr. lb. oz.dr. 
 
 73 11 1 21 94 2 11 8 9 
 
 39 17 2 12 47 3 17 11 10 
 
 Here 251b. a qr. 
 
 APPLICATION. 
 
 1. Bought 17cwt. 2qr. 141b. of sugar, of which I sold 
 9cwt. 3qr. 251b. ; how much remains unsold 1 
 
 Ans. 7cwt. 2qr. 171b. 
 
 2. Bought, at one time, 9cwt. 3qrs. 211b. 8oz. of iron, 
 and sold, next day, 8cwt. Iqr. 241b. 14oz. ; bought at 
 another time, 15cwt. 131b. 15oz. of the same kind, and sold 
 the same week, 15cwt. Iqr. ; what remains unsold of the 
 two parcels, 251b. a qr. ? Ans. Icwt. Iqr. lOlb. 9oz. 
 
 APOTHECARIES' WEIGHT. 
 lb. 1. 5- 9- gr. lb. f. 3- 9- gr. 
 
 81 7 2 11 69 10 4 1 12 
 
 47 2 4 2 15 30 11 1 2 17 
 
 APPLICATION. 
 
 Bought of an apothecary sundry articles, weighing 
 18ft. 5^ 2Q ; he compounded two parcels from them, 
 one of which weighed 7ft. 75. 19gr., and the other, 4ft. 
 45. l^. Igr. : how much was left uncompounded? 
 
 Ans. 2ft. 15. 
 
 CLOTH MEASURE. 
 
 Yd. qr. na. E. FL qr. na. E. E. qr. na. E. Fr. qr. na. 
 42 1 2 74 1 3 21 3 1 89 4 2 
 
 17 2 1 40 2 1 942 12 
 
 APPLICATION. 
 
 1. From a fashionable piece of cloth, which contained 
 52yds. 2na., a tailor was ordered to take 3 suits, each 
 6yds. 2qrs. ; how much remains of the piece ? 
 
 Ans. 32yds. 2qrs. 2na. 
 
58 COMPOUND SUBTRACTION. 
 
 2. Swapped with John Jones a piece of Irish linen 
 containing 36yds. Iqr. 3na. for a piece of Holland con- 
 taining 20E. Fl. 2qrs. Ina. : and swapped also with Seth 
 Sears a piece of sheeting measuring 41E. E. 4qrs. 2na. 
 for a piece of French cambric measuring 30E. Fr. 5qs. 
 3na. ; each agreed to pay me the balance, in quantity, in 
 jean : how many yards of jean must I receive from both ? 
 
 Ans. 26yds. 3qrs. Ina. 
 
 LONG MEASURE. 
 
 Deg. mi. fur. pol.ft. in. bar. Mi.fur.poL yd. ft* 
 81 30 4 24 12 6 1 74 3 16 4 1 
 
 29 41 5 13 14 12 48 5 37 2 2 
 
 APPLICATION. 
 
 1. Paul travelled three days, going 33mi. 5fur. 37pol. 
 2yd. 2ft. each day ; Amos set out with him, but travelled 
 only 28mi. 7fur. lyd. 2ft. each day ; at the end of the 
 third day, how far was Amos behind Paul ? 
 
 Ans. 14mi. 4fur. 31pol. 3yd. 
 
 2. The ship Sea-horse, bound to a port at 320 leagues' 
 distance, sailed 6 days at an average rate of 120mi. 18pol. 
 9in. every 24 hours ; at the end of the six days, how far 
 short of her destined port was she 1 
 
 Ans. 791eag. 2mi. 5fur. llpol. 12ft. 
 
 LAND OR SQUARE MEASURE. 
 
 Acre. roo. per. yd. ft. in. Ac. roo. per. 
 
 93 2 27 14 7 101 39 2 17 
 
 64 3 14 16^ 8 97 16 3 19 
 
 APPLICATION. 
 
 1. Rufus, David, Moses, and Robert owned, together, 
 542ac. Iroo. 34rods, 29^yds. 7ft. 142in. of land ; the 
 first had lOlaer. 3roo. 21rod. 4ft. 139in. the 2d, 99acr. 
 4fyd. ;Vthe 3d, 97acr. 24per. 8ft. 99in. ; how much then, 
 had the fourth ? 
 
 Ans. 244ac. Iroo. 29per. 23yds. 3ft. 48in. 
 
COMPOUND SUBTRACTION. 59 
 
 3. A man had 1000 acres of land, which he divided 
 among his three sons ; giving Abraham 156ac. 3roo. 14 
 rod. 27yds, 8ft. 125in. ; Isaac as much again ; and Jacob 
 the rest ; pray how much had Jacob ? 
 
 Ans. 529ac. Iroo. 85per. 6fyd. 57in. 
 
 CUBIC OR SOLID MEASURE. 
 
 T. rou. ft. inches. T. hewn. ft. Cord. ft. 
 
 64 37 1141 802 26 31 39 
 
 17 19 1400 204 31 9 107 
 
 APPLICATION. 
 
 1. Agreed with Noah Nott for 134 tons, 20ft. of round 
 timber T he has drawn 99tons, 39ft. 1701m. ; how much 
 more must he haul to complete the contract ? 
 
 Ans. 34tons, 20ft. 27in. 
 
 2. 1 bought 94tons, 25ft. 1600in. of hewn timber, of 
 which I have sold F. Francis 54tons, 45ft. 1709m. ; and 
 Saul Swift agrees to take the rest ; how much less will be 
 the quantity taken by Swift than that taken by Francis ? 
 
 Ans. 15tons, 16ft. 90in. 
 
 DRY MEASURE. 
 
 ChaL bus. pk. Bus. pic. gal. qt. pt. 
 
 103 17 1 62 1 2 1 
 
 94 31 2 15 2 3 1 
 
 APPLICATION. 
 
 1. A merchant contracted for 6000 bushels of wheat, 
 which was shipped on board of two vessels ; one arrived, 
 and brought 3000bush. 2pk. Igal. 3qt. Ipt. ; the other nev- 
 er came into port ; how much less was lost than was saved? 
 
 Ans. Ibus. Ipk. Igal. 3qts. 
 
 2. Having 199 chaldrons of coals, I sell George Green 
 99chal. 34 bus. Ipk., and Mark Mann the rest ; how much 
 less has Mark than George ? Ans. 32bus. 2pk. 
 
80 DECIMAL FRACTIONS. 
 
 mal point, in the sum, directly under the decimal points 
 of the numbers which have been added. 
 
 EXAMPLES. 
 
 124,6201 3741,21 
 
 5,92 374,646 
 
 17,1174 8,46 
 
 305,2165 52117,42 
 
 2,71 91,5 
 
 455,5840 
 
 In this first example, the sum is 455 integers, or pounds, 
 and yVVb- parts of a unit or pound ; or it is 455 units, and 
 5 tenth parts, 8 hundredth parts, and 4 thousandth parts 
 of a unit, or 1 ; the cipher at the right of the decimal 
 places does not affect the value of the other figures, and 
 it is, therefore, thrown away. 
 
 Hence we may observe, that decimals, and Federal 
 Money, are subject to one and the same law of notation, 
 and, consequently, of operation. 
 
 For since 1 dollar is the money unit, and a dime being 
 the tenth, a cent the hundredth, and a mill the thousandth 
 part of a dollar, it is evident that any number of dollars, 
 dimes, cents, and mills, is simply the expression of dol- 
 lars, and decimal parts of a dollar: thus, 11 dollars, 6 
 dimes, 5 cents=ll,65 or H T 6 ^dol. &c. 
 
 3. What is the sum of 276,+39,213+72014,9+417,+ 
 5032, and+2214,298 acres ? Ans. 79993,411 acres. 
 
 4. What is the sum of ,014+,9816+,32 + , 15914 + 
 ,72913+, 0047 gallons ? Ans. 2,20857gal. 
 
 5. What is the sum of 27,148+918,73+14016,+294304, 
 +7138, and +221,7 bushels ? Ans, 316625,578bus. 
 
 6. Add the following sums of dollars together, viz. 
 $12,34565+7,89l+2,34+14,+0011. 
 
 Ans. $36,57775 or $36, 5di. 7cts. 
 
 7. To 9,999999 miles add one millionth part of a mile-.. 
 
 Ans* 10rail.es*. 
 
DECIMAL FRACTIONS. 81 
 
 SUBTRACTION OF DECIMALS. 
 
 RULE. Set the less number under the greater in the 
 same manner as in addition ; then subtract as in whole 
 numbers, and place the decimal point in the remainder 
 directly under the other points. 
 
 EXAMPLES. 
 
 $. . 
 
 612,32 16,279 
 
 51,0942 8,0917 
 
 561,2258 
 
 4. From ,9173ft. subtract ,2138. Ans. ,7035 foot. 
 
 5. From $2,73 subtract $1,9185. Ans. $,8115. 
 
 6. Subtract 91,713acres from 407. Ans. 315,287ac. 
 
 7. What is the difference between 67tons and ,92 of a 
 ton ? Ans. 66,08 tons. 
 
 8. From 1 league subtract the millionth part of itself. 
 
 Ans. ,999999 league. 
 
 MULTIPLICATION OF DECIMALS. 
 
 RULE. Place the factors, (whether mixed numbers, or 
 pure decimals,) and multiply them, as in whole numbers ; 
 and from the product, towards the right hand, point off 
 as many figures for decimals as there are decimal places 
 in the factors. But if there be not so many figures in 
 the product, prefix ciphers' to supply the defect. 
 
 * 
 EXAMPLES. 
 
 1. 2. 
 
 21,41 yards. ,2616 ft. 
 
 25,9 shillings. ,154 doll. 
 
 19269 10464 
 
 10705 13080 
 
 4282 2616 
 
 554,519 shill. Ans. ,0402864 doll. Ans. 
 
62 COMPOUND MULTIPLICATION. 
 
 AVOIRDUPOIS WEIGHT. 
 
 Ton. cwt. qr. ft. ft. oz. dr. 
 
 1 17 3 23 21 11 15 
 
 5 6 
 
 Here 25ft. a qr. 
 
 APPLICATION. 
 
 1. What is the weight of 6 barrels of sugar, each 
 weighing Icwt. 3qrs. 20ft,. 1 Ans llcwt. 2qrs. 8ft. 
 
 2. What is the weight of 12 hogsheads of sugar, each 
 13cwt. 2qrs. 23ft., 25ft. a qr, 1 Ans. 164cwt. 3qrs. 1ft. 
 
 3. What is tlie weight of 6 chests of tea, each weigh- 
 ing 3cwt. 2qrs. 9ft., 25ft. a qr. ? Ans. 21cwt. 2qr. 41b. 
 
 CLOTH MEASURE. 
 
 Yds. qr. na. E.FL qr. na. E.E. qr. na. 
 313 91 a 1 61 4 3 
 
 789 
 
 APPLICATION. 
 
 1. What number of yards is in 8 pieces of broadcloth, 
 each 32yds. 3qrs. Ina. ? Ans. 260yds. 2qrs. 
 
 2. If 8 E. . 3qrs. 3na. of broadcloth will make a suit 
 of clothes; how much of the same cloth will make 12 
 similar suits 7 Ans. 105 . Eng. 
 
 LONG MEASURE. 
 
 Deg. mi. fur. pol. ft. in. bar. Mi. far.pol.yds. ft. 
 5 31 3 27 14 8 2 7 5 21 5 1 
 
 10 11 
 
 APPLICATION. 
 
 1. How far will a man travel in 7 days, if he go3lmi. 
 31 pol. 6ft. 6in. every day ? 
 
 Ans. 217mi. 6fur. I9pol. 1 2ft. 6in. 
 
 2. If in a race, a horse move 14ft. 7in. 2bar. at every 
 bound, and take 2 bounds in every second, what course 
 will he run over in 12 seconds ? Ans. 1 17yds. 4in. 
 
COMPOUND MULTIPLICATION. 63 
 
 LAND OR SQUARE MEASURE. 
 Ac. roo. per. yd. ft. in. Ac. roo. per. 
 
 15 3 19 23 7 72 39 1 37 
 
 8 9 
 
 APPLICATION. 
 
 1. In 9 fields each containing 14 acres, 1 rood, and 25 
 perches, how many acres ? Ans. 129ac. 2roo. 25per. 
 
 2. If a man divide his farm among his seven sons, and 
 give each 51 ac. 31 per. 8ft. how much does the farm con- 
 tain ? Ans. 358ac. Iroo. 17per. 6yd. 2ft. 
 
 CUBIC OR SOLID MEASURE. 
 T.rou. ft. in. T. hewn, ft. Cord. ft. 
 
 1 39 845 24 49 12 124 
 
 5 89 
 
 APPLICATION. 
 
 1 . Bought 5 boat loads of round timber, each of which 
 contained 3tons, 28ft. llllin. ; what is the whole quan- 
 tity 1 Ans. 18tons, 23ft. 371in. 
 
 2. In six parcels of wood, each containing 5cords, 96ft, 
 how many cords 1 Ans. 34cords. 
 
 DRY MEASURE. 
 
 Chal. bus. pic. Bus. pk. gal. qt. pt. 
 
 54 35 2 93121 
 
 6 7 
 
 APPLICATION. 
 
 \ . There were six wagons loaded with coals, each of 
 which contained Ichal. 8bus. 3pk. ; what was the total 
 quantity ? Ans. 7chal. 16bus, 2pk. 
 
 2. Nine loads of wheat were bought by a miller, each 
 of which contained 21bus. Ipk. Igal. Iqt. Ipt. ; what was 
 the total quantity ? Ans. 192bus. 3pk. Iqt. IpL 
 
64 COMPOUND MULTIPLICATTON. 
 
 WINE MEASURE. 
 
 Tun. lihd. gal. qt. Hhd. gal. qt. pt. gill 
 
 10 3 17 1 1 49 2 1 I 
 
 11 12 
 
 APPLICATION. 
 
 1. How much wine in 9 casks, each containing 41gal. 
 3qts^ Ipt. ? Ans. 376gals. 3qts. Ipt. 
 
 "2. How much cider may be put into eight casks, each 
 of which will hold 104gals. 2qts. Ipt. 3gills ? 
 
 - Ans. 837f gals. 
 
 TIME. 
 
 Yr. m. w. d. Ji. m. s. Yr. days. h. m. 
 
 4 12 2 5 10 10 10 1 224 5 40 
 
 12 8 
 
 APPLICATION. 
 
 If a solar year contain just 365 days, 5 hours, 48 min- 
 utes, and 48 seconds, what time is equal to 7 solar years ? 
 Ans. 2556days, 16h. 4 1m. 36s. 
 
 CASE II. If the multiplier exceed 12, multiply suc- 
 cessively by its component parts, as in Case II in sim- 
 ple Multiplication. 
 
 EXAMPLES. 
 
 1. What will &lyds. of calico come to, at 2s. 7d. per 
 yard ? 
 
 3x7=21 
 
 18 4J price of 7 yards. 
 3 
 
 Ans. ,2 15 1J- price of 21 yards. 
 
 2. What will 18 yards come to, at 1 7s. %d. per 
 yard 1 Ans, <24 9s. 9d. 
 
 3. What will 36 pair of shoes come to, at 13s. 4d. per 
 pair? Ans. 
 
COMPOUND MULTIPLICATION. 65 
 
 4. What will 49 yards of broadcloth cost, at 17s. 
 per yard ? Ans. ,43 Os. 
 
 5. A gentleman is possessed of l doz. silver spoons, 
 each weighing 2oz. I5dwt. llgr. ; 2 doz. teaspoons, each 
 Wdwt. I4gr. ; and two silver tankards, each21oz. \5dwt. ; 
 pray what is the weight of the whole ? 
 
 Ans. 8/6. lOoz. 2dwt. 6gr. 
 
 6. What is the weight of 42 tubs of butter, each 16/6. 
 Uoz. 126/r., 251b. a qr. ? Ans. 7cwt. 10/6. lloz. Sdr. 
 
 7. How many yards in 81 pieces of cloth, each 7yds. 
 3#rs. Ina. 1 Ans. 632y*/s. 3qrs. \na. 
 
 8. How many bushels in 63 casks, each containing 
 46ws. 3pks. Igal. 1 Ans. 3076ws. IgaL 
 
 CASE III. If the multiplier be not a composite num- 
 ber, find the nearest to it, either greater or less ; multi- 
 ply by the component parts as before, and for the odd 
 parts add or subtract as the case requires. 
 
 EXAMPLES. 
 
 1. What will 65 yards of cloth come to, at <! 14s. 
 i. yer yard ? 
 
 1 14 6 
 8 
 
 8x8=64 13 16 4 price of eight yards. 
 
 8 
 
 110 10 8 price of 64 yards. 
 4) 1 14 6J price of 1 yard. 
 8 7 price of yard.+ 
 
 Ans. ,112 13 10 price 65 % yards. 
 2. What will 76 yards cost, at 14s. 9d. per yard ? 
 
 <0 14 94 
 7x11=771=76 II 
 
 8 2 8 price of 11 yards. 
 
 7 
 
 56 18 11 price of 77 yardg. 
 Subtract 14 9 price of 1 yard. 
 
 Ans. ,56 4 2 price of 76 yards. 
 F2 
 
66 COMPOUND DIVISION. 
 
 3. What will 183 gallons wine cost, at 7s. 5d. ? 
 10x10=100 
 
 10xS=80 
 
 3=3 Then 100+80+3=183. 
 
 Ans. ,67 17s. 3d 
 
 4. What will 600 yards of cloth cost, at <! 2s. 7d. 1 
 
 10x10x6=600 Ans. .678 15s. Odf. 
 
 5. If a man travel 46m?". 7fur. 30pol 3yds. 2ft. every 
 day, for 57 J days, how far will he go ? 
 
 Ans. 2700mi. Cyfur. 23poL lyd. 2ft. 6in. 
 
 6. If a farm consist of 43 lots, and each lot contain 
 3ac. 3r0o. 3per. 3yds. 3ft. 3m. ; how large is the farm ? 
 
 Ans. 162c. I3per. 22yds. 3ft. 129m. 
 
 7. In 51 loads of bark, each 1 cord, 26 feet, how many 
 cords ? Ans. 61cor. 46/fc. or 61e0r. 2ft. bark meas. 
 
 8. If a cargo of wine consist of 122 casks, and each 
 cask contain 51gal. 3qts. Ipt. how many tuns in all ? 
 
 Ans. 25tun, 28gal. 3qts. 
 
 9. If a person waste Ihour, 32min. 41sec. every day 
 for 38 years, how much time does he lose in the whole 
 ^period, allowing 365 days to the year ? 
 
 Ans. 2yrs. 162da. 19h. 58rn. 
 
 COMPOUND DIVISION. 
 
 j+ 
 
 COMPOUND DIVISION teaches to find how often one num- 
 ber is contained in another of different denominations. 
 
 RULE. Place the numbers as in Simple Division. Be- 
 gin at the left hand, and divide each denomination by the 
 divisor, setting the quotients under their respective divi- 
 dends ; but if there be a remainder in dividing any of the 
 denominations except the lowest, find how many of the 
 next lower denomination it is equal to ; and add it to the 
 number, if any, which was in this denomination before ; 
 divide this sum as usual, and thus proceed, until the 
 whole is finished. 
 
 The method of proof is the same as in Simple Division. 
 
COMPOUND DIVISION. 67 
 
 CASE I. EXAMPLES. 
 
 MONEY, 
 
 s. d. s. d. qr. s. fl. 
 
 8)39 11 6 6)22 942 5)71 13 8f 
 
 4 18 1U+ 
 
 APPL1CATIOJV. 
 
 1. Divide 67 10s. 9d. by 7. Ans. 9 12s. 
 
 2. Divide S equally among 6 persons. 
 
 Ans. 1 6s. 8d. 
 
 TROY WEIGHT 
 
 ft. oz. dwt. ft. oz. dwt. 
 
 4)13 1 15 6)1 
 
 APPLICATION. 
 
 If 2ft. 9oz. 5dwt. 12gr. of silver be wrought into a 
 dozen spoons, what will each weigh ? 
 
 Ans. 2oz. 15dwt. llgr. 
 
 AVOIRDUPOIS WEIGHT. 
 
 Cict. qr. ft Ton. act. qr, 
 
 8)75 1 12 8)8 
 
 Here 25ffo a qr. 
 
 APPLICATION. 
 
 1. Divide 13cwt. Iqr. 12ft. 6oz. lOdr. by 11. 
 
 Ans. Icwt. 24ft. 9 T 7 T dr. 
 
 2. Divide 17cwt. Iqr. of sugar equally among 6 persons, 
 at 25ft a qr. ? Ans. 2cwt. 3qrs. 
 
 CLOTH MEASURE. 
 
 Yd. qr. na. E.E. qr. na. 
 
 5)31 23 9)1 00 
 
68 COMPOUND DIVISION. 
 
 K 
 
 APPLICATION. 
 
 1. If 8 equal pieces of cloth contain 260yds. 2qrs. what 
 does each piece contain 1 Ans. 32yds. 2qr. Ina. 
 
 2. If 105 E. Eng. will make 12 suits of clothes, what 
 does it take for one suit ? Ans. 8E. E. 3qr. 3na. 
 
 LONG MEASURE. 
 
 Deg. mi. fur. pol. ft. in. bar. Mi fur. pol. yd. ft. 
 10)6 31 3 28 14 8 2 11)7 5 21 5 1 
 
 APPLICATION. 
 
 1. If a man travel 217m iles, 5fur, I9pol. 12ft. 6in. in 
 7 days ; how far does he go a day 1 
 
 Ans. 31 mi. 31pol. 6ft. 6in. 
 
 2. If, in a race, a horse go over a course of 117yds. 4in. 
 in 12 seconds, how far does he move in one second ? 
 
 Ans. 9yds. 2ft. 3in. Iban 
 
 LAND OR SQUARE MEASURE. 
 
 Ac. roo. per. yd. ft. in. Ac. roo. per. 
 
 8)15 3 20 87 72 9)39 1 37 
 
 APPLICATION. 
 
 1. If nine fields of equal extent, contain 129ac. 2roo. 
 25per. what does one of them measure ? 
 
 Ans. 14ac. Iroo. 25per. 
 
 2. If a man divide his farm of 35Sac. Iroo. 17per. 6yd. 
 2ft. in equal portions between seven sons, what does each 
 have 1 Ans. 51ac. 3lper. Sft. 
 
 CUBIC OR SOLID MEASURE. 
 
 T.rou.ft. in. T. hew. ft. Cord. ft. 
 
 5)1 39 845 8)24 49 9)12 124 
 
COMPOUND DIVISION. 69 
 
 APPLICATION. 
 
 1. Bought I8tons, 23ft. 371in. of round timber, which 
 was in five boats, and each contained a like quantity; 
 how much did one boat contain 1 
 
 Ans. 3tons, 28ft. llllin. 
 
 2. In six equal parcels of wood I have 34^cords ; what 
 is in each parcel. Ans. 5cords, 96ft. 
 
 DRY MEASURE. 
 
 Chal bus. pk. Bus. pic. gal qt. pt. 
 
 6)54 35 2 7)9 3 1 2 I 
 
 APPLICATION. 
 
 1. If in nine equal loads of corn there be 192bus. 3pk. 
 Iqt. Ipt., what is there in one load 1 
 
 Ans. 21bus. Ipk. Igal. Iqt. Ipt. 
 
 2. Six wagons equally loaded drew to market 7chal. 
 16bus. 2pk. of coals ; how much did one bring ? 
 
 Ans. Ichal. 8bus. 3pk. 
 
 WINE MEASURE. 
 
 Tuns.hhd. gal qt. Hhd. gal. qt. pt. gill. 
 
 11)19 3 17 1 12)1 49 2 1 1 
 
 APPLICATION. 
 
 1. In nine equal casks, I have 376gal. 3qt. Ipt. of 
 wine ; how much is in one cask ? 
 
 Ans. 41gaL 3qts. Ipt. 
 
 2. I have put into 8 hogsheads 837Jgal. of cider, each 
 being filled alike ; how much is in each cask ? 
 
 Ans. 104gal. 2qts. Ipt. 3gills. 
 
 TIME. 
 
 Yr. m. w. d. 7i. m. s. Yr. d. h. m. 
 
 12)4 12 2 5 10 10 10 5)4 224 5 40 
 
70 COMPOUND DIVISION. 
 
 APPLICATION. 
 
 If in seven solar years, there be just 2556days, 16b. 
 41m. 36s., what is the length of one solar year ? 
 
 Ans. 365days, 5h. 48m. 48s. 
 
 CASE II. If the divisor exceed 12, divide continually 
 by its component parts, as in Simple Division, CASE III. 
 
 EXAMPLES. 
 
 1. Divide <37 16s. equally among 24 men. 
 
 '. s. d. 
 6)37 16 
 
 4)6 6 
 1 11 6 Ans. 
 
 2. If 20 gallons of wine cost 7 5s. lOd. what is it 
 per gallon ? Ans. 7s. 3d. 
 
 3. Bought 3doz. of silver spoons, which, together, 
 weighed 4]fo. 8oz. 12gr. : how much silver did each spoon 
 contain ? Ans. 3oz. 4dwt. llgr. 
 
 4. Divide a hhd. of sugar, weight 12cwt. 3qrs. 7|fc 
 equally among 16 men, 25ft. a qr. Ans. 3qrs. 5Jfo. 2oz. 
 
 5. Divide 43yds. Iqr. Ina. of crape among 33 persons. 
 
 Ans. lyd. Iqr. Ina. 
 
 6. If a person travel 171eag. Imi. 4fur. 21pol. in 21 
 hours, what was the average distance an hour t 
 
 Ans. 2mi. 4fur. Ipol. 
 
 7. Divide 1000 acres of land equally among 99 per- 
 sons. Ans. lOac. 16fper. 
 
 8. Divide 500 cords of bark in equal parts among 108 
 persons. Ans. 4cor. 80 T 6 Q 4 ft. or 4cor. S^yft. bark meas. 
 
 9. Divide 168bus. Ipk. IgaL 2qts. of corn equally 
 among 35 persons. Ans. 4bus. 3pk. 2qt. 
 
 10. Divide 4^gal. of wine equally among 144 soldiers. 
 
 Ans. Igill each. 
 
 CASE III. If the divisor be not a composite number, 
 divide as in long division. 
 
COMPOUND DIVISION. 71 
 
 EXAMPLES. 
 ' 1. Divide ,391 17s. 6d. equally among 46 men. 
 
 s. d. s. d. 
 46)391 17 6(8 10 4 Ans. 
 368 
 
 23 
 20 
 
 46)477(105. 
 46 
 
 17 
 12 
 
 46)210(4d. 
 184 
 
 26 
 4 
 
 46)105(2^5. 
 92 
 
 13 
 
 2. If 263 bushels of wheat cost ^86 11s. 5d. what is 
 it per bushel 1 
 
 s. d. 
 
 263)86 11 5(<0 6s. 7d. Ans. 
 20 
 
 263)1731 (6s. <fcc. 
 
 3. Divide 27tons, 13cwt. 2qrs. of iron equally among 
 34 men, 251b. a qr. 
 
 T. cwt. qr.T. cwt. qr. ft. 
 34)27 13 2(0 16 1 2 and 32 rem. or ff. 
 20 
 
 34)553(16cwt. <fcc. Ans. 16cwt. Iqr. 21b. 
 
 4. If 461b. of indigo cost ^53 10s. 6d., what i it per 
 pound ? Ans. \ 35. 
 
72 COMPOUND DIVISION. 
 
 5. If 263 bushels of wheat cost $287, 97cts. 3mills, 
 what is it per bushel ? Ans. $1, 9cts. 4mills..{- 
 
 6. If 37 thousand of boards come to $203,50cts., what 
 is one thousand worth ? Ans. $ 5,50cts. 
 
 7. Divide 120 months, 2w. 3d. 5h. 20m. by 111. 
 
 Ans. Imo. 2d. lOh. l2 T 6 T 8 T m. 
 
 8. A privateer takes a prize worth $12465, of which 
 the owner takes one-half, the officers one-fourth, and the 
 remainder is equally divided among the sailors, who are 
 125 in number ; how much is each sailor's part ? 
 
 Ans. $24,93cts. 
 
 9. Three merchants, A, B, and C, have a ship in com- 
 pany ; A has f, B f , and C ; they have received for 
 freight ,228 16s. 8d. ; it is required to divide it among 
 the owners, according to their respective shares ; pray, 
 can you do it ? . ( A's share <143 5d. B's ,57 
 
 s '\ 4s. 2d. C's ^28 12s. Id. 
 
 10. A privateer having taken a prize worth $6850, it 
 is divided into one hundred shares ; the captain takes 1 1 ; 
 2 lieutenants, each 5 ; 12 midshipmen, each 2 ; and the 
 remainder is to be equally divided among the sailors, who 
 are 105 in number; pray, can you settle the matter ? 
 
 . i Captain's share $753,50cts. ; a lieut's $342,50cts. ; 
 s * \ a midshipman's $137; and a sailor's $35,88 T V\cts. 
 
 11. Divide the sum of 50 eagles, 50 dollars, 50 dimes, 
 50 cents, and 60 mills among 17 men ; and give the first 
 12 cents more than the second, the second 12 cents more 
 than the third, arid so on to the last; what will the sev- 
 enteenth man's share be 1 Ans. $31,72cts. 
 
 NOTE. It is customary among appraisers of property, 
 arbitrators or referees, and, in some cases, the method is 
 resorted to even by juries, where they cannot agree in 
 their estimates and verdicts, to take the amount of the 
 sums which they severally agree to award, and divide that 
 amount by the number of which they consist ; the quo- 
 tient is their average judgment, which goes for their esti- 
 mate, decision, or verdict. 
 
 EXAMPLES. 
 
 1. In appraising a certain property, A called the value 
 $100, B, $140, C, $80, and D, $150 ; but as they could 
 
DUODECIMALS. 
 
 73 
 
 not agree in their estimates, they determined to take their 
 average judgment, and let that be the value ; what was 
 the value in that case ? 
 
 A $ 100 
 
 B 140 
 
 C 80 
 
 D 150 
 
 4 )470 
 
 $117,50c. Ans. 
 
 2. Four persons appraising a building, A valued it at 
 $550,50c., B, at $480,50c. , at $590,50c. and D, at 
 $600 ; what was the mean value? Ans. $555,37^c. 
 
 3. Seth Strong, Luke Locke, Mark Mills, Charles 
 Church, and John Jones, were appointed to appraise the 
 ship Ocean ; Strong's estimate was $5500, Locke's $7000, 
 Mill's $6666, Church's $8444, and Jones's, $5000; 
 what was the mean judgment ? Ans. $6522. 
 
 DUODECIMALS. 
 
 DUODECIMALS chiefly regard feet and inches. They 
 are so called, because they decrease by twelve from the 
 place of feet towards the right hand. 
 
 Inches are sometimes called primes, and marked thus 
 (') ; the next division is called parts or seconds, and 
 marked (*) ; the next thirds, and marked ( m ) ; &c. 
 
 MULTIPLICATION OF DUODECIMALS. 
 
 RULE. Under the multiplicand write the corresponding 
 denomination of the multiplier. Multiply each term in 
 the multiplicand, beginning at the lowest, by the highest 
 denomination in the multiplier ; and write each result 
 under its respective term ; observing to carry a unit for 
 every 12 from each lower place to its next higher. 
 
 In the same manner multiply all the multiplicand by 
 the next highest denomination in the multiplier; and set 
 G 
 
74 DUODECIMALS. 
 
 the result of each term removed one place to the right 
 hand of those in the multiplicand. 
 
 Proceed in like manner with the remaining denomina- 
 tions, and the sum of all the lines will be the product re- 
 quired. 
 
 EXAMPLES. 
 
 1. Multiply 4 feet 2 inches by 3 feet 5 inches. 
 
 Ft. ' 
 
 The 2 in the product 4 2 
 is not 2 inches, but 235 
 
 twelfths of a square foot, 
 
 or 24 inches, &c. 12 6 
 
 1 8 10" 
 
 14 2 10 Ans. 2ft. 2' 10". 
 
 2. Multiply 10 feet 11 inches by 7 inches. 
 
 10 11 
 
 7 
 
 645 Ans. 6ft. 4' 5". 
 
 3. What is the content of a bale 6 feet 5' long ; 4 feet 
 3' high, and 3 feet 10' wide ? 
 
 Ft. ' 
 6 5 
 4 3 
 
 25 8 H 
 173 
 
 27 3 3 
 3 10 
 
 81 9 9 m 
 
 22 8 8 6 
 
 104 6 5 "6 Ans. 104ft. 6' 5" 6 W . 
 
VULGAR FRACTIONS. 75 
 
 4. What is the content of a marble slab 4ft. ?' S" wide 
 and 5ft. 6' long? Ans. 25ft. 6' 2". 
 
 5. Multiply 7ft. 8' 6" by 10ft. 4' 5". 
 
 Ans. 79ft. 11' 0" 6'" 6"". 
 
 6. Multiply 44ft. 2' 9" 2'" 4"" by 2ft. 10' 3". 
 
 Ans. 126ft. 2' 10" 8'" 10"" 11'"". 
 
 7. How many square feet in a board 25 feet 6 inches 
 long, and 1 foot inches wide 1 
 
 Ans. 31 feet 10 inches. 
 
 8. How many cubic feet in a stick of timber 12 feet 
 10' long, 1 foot 7' wide, and 1 foot 9' thick 1 
 
 Ans. 35ft. 6' 8 V 6'". 
 
 9. How many cubic feet of wood in a load 7 feet 10' 
 long, 3 feet 11' wide and 3 feet 6' high 1 
 
 Ans. 107ft. 4' 7". 
 
 10. The length of a room being 20ft., its width 14ft. 6', 
 and height 1 Oft. 4' ; how many yards of painting are in it, 
 deducting a fire-place of 4ft. by 4ft. 4', and two windows, 
 each 6ft. by 3ft. 2' ? Ans. 73^-yards. 
 
 11. Required the solid contents of a wall 53ft. 6' long, 
 10ft. 3' high, and 2ft. thick. Ans. 1096ft. 9'. 
 
 FRACTIONS. 
 
 FRACTIONS, or broken numbers, are expressions for any t 
 assignable parts of a unit, or whole number, and, gene- 
 rally, are of two kinds, viz. 
 
 VULGAR AND DECIMAL. 
 
 A Vulgar Fraction is represented by two numbers, 
 placed one above the other, with a little line drawn be- 
 tween them ; thus, , |, &c. signify three-fourths, five- 
 eighths, &c. 
 
 The figure above the line, is called the numerator, and 
 the one below the line the denominator. 
 
 ( 5 Numerator. 
 Thus J 
 
 ( 8 Denominator. 
 The denominator (which is the divisor, in division,) 
 
76 VULGAR FRACTIONS. 
 
 shows how many parts the unit or integer is divided into ; 
 and the numerator (which is the remainder after division,) 
 shows how many of those parts are meant by the fraction : 
 A fraction is said to be in its least or lowest terms, 
 when it is expressed by the least numbers possible ; thus 
 | when reduced to its lowest terms, will be ^, and T 9 ^ when 
 expressed by the least numbers possible, will be f ; and 
 so of any others. 
 
 PROBLEM I. , 
 
 To abbreviate or reduce fractions to their lowest terms. 
 
 RULE. Divide the terms of the given fraction by any 
 number which will divide them without leaving a remain- 
 der, and the quotients divide again in like manner ; and 
 so on till it appears that there is no number greater than 
 1, which will any longer divide them \ and then the frac- 
 tion will be in its lowest terms. 
 
 EXAMPLES. 
 
 1. Reduce f-|- to its lowest terms. 
 
 =f the answer. 
 
 2. Reduce -J-f f to its lowest terms. Ans. ^. 
 
 3. Reduce f ^f to its lowest terms. Ans. \. 
 
 4. Reduce /Y\- to its lowest terms. Ans. ^. 
 
 5. Abbreviate ff as much as possible. Ans. \^. 
 
 0. Bring ff^ to"the least terms possible. Ans. ff . 
 
 7. Reduce ^ff to its lowest terms. Ans. f. 
 
 8. Abridge % as low as you can. Ans. . 
 
 9. Put \\ into its least terms. Ans. %. 
 10. Let I iff be expressed in its least terms. Ans. f. 
 
 PROBLEM 2. 
 
 To find the value of a fraction in the known parts of 
 
 the integer, as of coin, weight, measure, fyc. 
 RULE. Multiply the numerator by the common parts 
 of the integer, and divide by the denominator, and the 
 several quotients set in one line, will show the answer. 
 
 EXAMPLES. 
 
 1. What is the value of f of a pound sterling 1 
 
VULGAR FRACTIONS. 77 
 
 Numer. 2 
 
 20 shillings in a pound. 
 
 Denom. 3)40(l3s. 4d. the Ans. 
 3 
 
 10 
 9 
 
 1 
 12 pence in a shilling. 
 
 3)12(4d. 
 12 
 
 2. What is the value of % of a pound sterling? 
 
 Ans. 18s. 5d. 2 T 2 oqrs. 
 
 3. Reduce f of a shilling to its proper quantity. 
 
 Ans. 4d. 3^qrs. 
 
 4. What is the value of f of 12s. 6d. 7 Ans. 4s. 
 
 5. What is the value of ff of a pound Troy 7 
 
 Aris. 9oz. 
 
 6. How much are T 9 T of a hundred weight ? 
 
 Ans. 3qrs. 7fc. 10 T 2 T oz. 
 
 7. What is the value of f of a mile ? 
 
 Ans. 6fur. 26pol. lift. 
 
 8. How much are of a cwt., at 25fb to a qr. ? 
 
 Ans. 3qrs. Sffe. 12oz. 7|dr. 
 
 9. Show the proper quantity of f of an ell English I 
 
 Ans. 2qrs. 3jna. 
 
 10. How much are % of a hhd. of wine ? 
 
 Ans. 54 gallons. 
 
 11. What is the value of T % of a day I 
 
 Ans. iGh. 36m. 55 T 5 TrS. 
 
 12. What is the value of of a dollar 1 Ans. 80cts. 
 
 PROBLEM 3* 
 
 To reduce any given quantity to the fraction of any 
 greater denomination of the same kind. 
 
 RULE. Reduce the given quantity to the lowest term 
 mentioned, (as in reduction of Money r Weights,, 
 G2 
 
78 VULGAR FRACTIONS. 
 
 nres, &c.) for a numerator ; then reduce the integral part 
 to the same term, for a denominator 1 place the former 
 over the latter, and they will be the fraction required, 
 which, if it be not in its lowest terms, must be reduced 
 by Problem 1. 
 
 EXAMPLES. 
 
 1. Reduce 13s. 6d. 2qrs. to the fraction of a pound* 
 
 5. d. qrs. 
 
 20 shill. integral part. 136 2 given sum. 
 12 12 
 
 240 162 
 
 4 4 
 
 960 Denominator,. 650 Numerator. 
 
 Ans. = 
 
 2. What part of a hundred weight are 3qrs. 
 
 Ans, T ^= 
 
 3. What part of a yard are 3qrs, 3n.a. T Ans. | 
 
 4. What part of a pound sterling are 13s. 4d. 
 
 Ans. 
 
 5. What part of a civil year are 3 weeks, 4 days 7 
 
 Ans. 3 2 g 5 ? = T 
 
 6. What part of a mile are 6fur. 26pol. 3yds. 2f.. 
 
 fur. pal. yds. ft. feet. 
 6 26 3 2==44QO Num. 
 
 a mile =5280 Denom. Ans. 
 
 7. Reduce 7oz. 4dwt. to the fraction of a }fr. Troy. 
 
 Ans. f . 
 
 8. What part of am acre are 2 roods, 20 perches 1 
 
 Ans. . 
 
 9. Reduce 54 gallons to the fraction of a hhd. of wine. 
 
 Ans. -|. 
 
 10. What part of a hogshead of wine are 9 gallons ? 
 
 Ans. ^. 
 
 11. What part of a pound Troy are lOoz. lOdwt. lOgr. 1 
 
 Ans. &%. 
 
 12. What part of a dollar are 80cts. T Ans. |-. 
 
DECIMAL FRACTIONS. 7*J 
 
 DECIMAL FRACTIONS. 
 
 A DECIMAL is a fraction whose denominator is a unit, 
 with as many ciphers annexed to it as the numerator has 
 places ; and is usually expressed by writing the numer- 
 ator only with a point before it called the separatrix. 
 Thus T 5 <y, T 2 <p T % 3 o 6 <T> are decimal fractions, and are ex-- 
 pressed by ,5 ,25 ,236 respectively. 
 
 The place of a figure in decimals, as in whole numbers, 
 determines its relative value : That in the first place next 
 the separatrix is 10th parts; that in the second, 100th 
 parts, &c. decreasing in the same tenfold proportion to 
 the right hand, as whole numbers increase decimally from 
 units to the left hand. 
 
 Ciphers placed at the right hand of decimals, make na 
 alteration in their value ; for ,5 ,50 ,500, &c. are decimals 
 of the same vajue, being each equal to ^ ; but. if placed 
 at the left hand, the value of the fraction is decreased in 
 a tenfold proportion for every cipher prefixed ; thus ,5 
 ,05 ,005, &<c. are 5 tenth parts, 5 hundredth parts, and 5 
 thousandth parts respectively. In the following Table, 
 the doctrine is exemplified at large. 
 
 NUMERATION TABLE, OR SCALE OF NOTA^ 
 TION, 
 
 S 
 
 7 
 
 03 
 
 o 
 G 
 
 3 
 O 
 
 EH 
 
 M 
 5 
 
 a 
 
 03 
 
 H 
 4 
 
 a 
 
 B 
 3 
 
 a .ti 
 
 <U 3 
 
 03 
 P. 
 
 s 
 
 CD 
 
 2 1, 
 
 as 
 
 3 
 
 M 
 3 
 
 03 
 OH 
 
 3 
 C 
 
 H 
 4 
 
 3 
 03 
 CO 
 
 3 
 O' 
 
 H 
 XI 
 5 
 
 03 
 
 P- 
 
 3 
 O 
 
 H 
 
 -* 
 03 
 
 O 
 
 6 7 
 
 WHOLE NUMBERS. 
 
 DECIMALS. 
 
 ADDITION OF DECIMALS. 
 
 RULE. Set the numbers so that the decimal points 
 may stand directly under each other, then add as in whole 
 s, carrying one for every ten, and place the deci- 
 
80 DECIMAL FRACTIONS. 
 
 mal point, in the sum, directly under the decimal points 
 of the numbers which have been added. 
 
 EXAMPLES. 
 
 $. 
 
 124,6201 3741,21 
 
 5,92 374,646 
 
 17,1174 8,46 
 
 305,2165 52117,42 
 
 2,71 91,5 
 
 455,5840 
 
 In this first example, the sum is 455 integers, or pounds, 
 and ^0% parts of a unit or pound ; or it is 455 units, and 
 5 tenth parts, 8 hundredth parts, and 4 thousandth parts 
 of a unit, or 1 ; the cipher at the right of the decimal 
 places does not affect the value of the other figures, and 
 it is, therefore, thrown away. 
 
 Hence we may observe, that decimals, and Federal 
 Money, are subject to one and the same law of notation, 
 and, consequently, of operation. 
 
 For since 1 dollar is the money unit, and a dime being 
 the tenth, a cent the hundredth, and a mill the thousandth 
 part of a dollar, it is evident that any number of dollars, 
 dimes, cents, and mills, is simply the expression of dol- 
 lars, and decimal parts of a dollar: thus, 11 dollars, 6 
 dimes, 5 cents=ll,65 or H T ^dol. &c. 
 
 3. What is the sum of 276,+39,213+72014,9+417,+ 
 5032, and-f 2214,298 acres ? Ans. 79993,411 acres. 
 
 4. What is the sum of ,014 + ,9816+,32 + , 15914 + 
 ,72913+, 0047 gallons ? Ans. 2,20857gal. 
 
 5. What is the sum of 27,148+918,73+14016,+294304, 
 +7138, and +221,7 bushels ? Ans, 316625,578bus. 
 
 6. Add the following sums of dollars together, viz. 
 $12,34565+7,891 +2,34+ 14,-fOOll. 
 
 Ans. $36,57775 or $36, 5di. 7cts. 7^mills. 
 
 7. To 9,999999 miles add one millionth part of a mile* 
 
DECIMAL FRACTIONS. 81 
 
 SUBTRACTION OF DECIMALS. 
 
 RULE. Set the less number under the greater in the 
 same manner as in addition ; then subtract as in whole 
 numbers, and place the decimal point in the remainder 
 directly under the other points. 
 
 EXAMPLES. 
 
 $. . yds. 
 
 612,32 16,279 37, 
 51,0942 8,0917 2,41 
 
 561,2258 
 
 4. From ,9173ft. subtract ,2138. Ans. ,7035 foot. 
 
 5. From $2,73 subtract $1,9185. Ans. $,8115. 
 
 6. Subtract 91,7 lucres from 407. Ans. 315,287ac. 
 
 7. What is the difference between 67tons and ,92 of a 
 ton ? Ans. 66,08 tons. 
 
 8. From 1 league subtract the millionth part of itself. 
 
 Ans. ,999999 league. 
 
 MULTIPLICATION OF DECIMALS. 
 
 RULE. Place the factors, (whether mixed numbers, or 
 pure decimals,) and multiply them, as in whole numbers ; 
 and from the product, towards the right hand, point off 
 as many figures for decimals as there are decimal places 
 in the factors. But if there be not so many figures in 
 the product, prefix ciphers' to supply the defect. 
 
 v 
 
 EXAMPLES. 
 
 1. 2. 
 
 21,41 yards. ,2616ft- 
 
 25,9 shillings. ,154 doll. 
 
 19269 10464 
 
 10705 13080 
 
 4282 2616 
 
 554,519 shill. Ans. ,0402864 doll. Ans. 
 
82 DECIMAL FRACTIONS. 
 
 3. Multiply 31,72rods by 65,3. Prod. 2071,316rods. 
 
 4. Multiply ,62foot by ,04. Prod. ,0248foot. 
 
 5. Multiply 51,6yards by 21. Prod. 1083,6yards. 
 
 6. Multiply ,051, by ,0091. Prod. ,0004641. 
 7 What cost 6,21 yards of cloth, at 2dols. 32cts. 5 
 
 mills per yard ? Ans. $14, 4d. 3c. 8^/wm. 
 
 8. Multiply 7,02 dollars by 5,27 dollars. 
 
 Ans. 36,9954dols. or $36,99cts. 5 T \m. 
 
 9. Multiply 41dols. 25cts. by 120dols. Ans. $4950. 
 
 10. Multiply 3dols. 45cts. by 16cts. 
 
 Ans. ,5520 or 55cts. 2m. 
 
 11. Multiply 65cts. by ,09 or 9cts. 
 
 Ans. ,0585=5cts. 8m. 
 
 12. Multiply lOdols. by lOcts. Ans. $1. 
 
 13. Multiply 341,45dols. by ,007 or 7 mills. 
 
 Ans. $2,39+. 
 
 NOTE. To multiply Decimal Fractions by 10, 100, 
 1000, &c. is only to remove the separatrix so many pla- 
 ces towards the right hand, as there are ciphers in the 
 multiplier. 
 
 EXAMPLES. 
 
 Multiply $64,674 by $10. Ans. $646,74. 
 
 Multiply $3,2158 by 1000 cords. Ans. $3215,8. 
 
 DIVISION OF DECIMALS. 
 
 RULE. Divide as in whole numbers ; and observe the 
 following rules for pointing off in the quotient. 
 
 1. Point off for decimals In the quotient so many fig- 
 ures, as the decimal places in the dividend exceed those 
 in the divisor. 
 
 2. If the figures in the quotient are not so many as the 
 rule requires, supply the defect by prefixing ciphers. 
 
 3. If the decimal places in the divisor be more than 
 those in the dividend, add ciphers as decimals to the divi- 
 dend, until the number of decimals in the dividend be, 
 equal to those in the divisor, and the quotient will be in- 
 tegers until all these decimals are used. And in case of 
 a remainder, after all the figures of the dividend are used, 
 and more figures are wanted in the quotient, annex ci- 
 
DECIMAL FRACTIONS. 83 
 
 pliers to the remainder, to continue the division as far as 
 necessary. 
 
 4. The first figure of the quotient will possess the same 
 place of integers or decimals, as that figure of the divi- 
 dend which stands over the unit's place of the first pro- 
 duct. 
 
 EXAMPLES. 
 
 1. Divide 3424,6056 by 43,6. 
 
 Divisor. Dividend. Quotient. 
 43,6)3424,6056(78,546 
 
 3052 Or, What is potash per 
 
 ton, when 43 tons, 12cwt. 
 
 3726 cost $3424,6056 ? 
 
 3488 Ans. $78,54c. 6m. 
 
 2380 
 
 2180 
 
 2005 
 1744 
 
 2616 
 2616 
 
 2. Divide 761,2 miles by 2,1942 weeks. 
 
 2,1942)761,2000(346,91 miles, and 10078 rem. 
 
 3. Divide 7,735 rods by 3,25 Ans. 2,38 rods. 
 
 4. Divide 3877875 by ,675. Ans. 5745000. 
 
 5. Divide 1835,78 tons bjt 7,48. Ans. 245,42tons.+ 
 
 6. Divide ,55736 cord by 48. Ans. ,01161cord.+ 
 
 7. Divide 7,13 acres by 18. Ans. ,396acre.+ 
 
 8. Divide 246,1476dols. by 603,25dols. 
 
 Ans. ,40736dol.+ 
 
 9. Divide 186513,239dols. by 304,81dols. 
 
 Ans. 6il,9dols.+ 
 
 10. Divide l,28dols. by 8,31 dols. 
 
 Ans. ,154=l5cts. 4m.+ 
 
 11. Divide 56cts. by Idol. 12cts. Ans.Sdimes or 50cts. 
 
 12. Divide 1 dollar by 12 cents. 
 
 Ans. 8,333dols. or, $8, 33cts. 3m. 
 
84 DECIMAL FRACTIONS. 
 
 13. If 2lf or 21,75 yards of cloth cost 34,517dols., 
 what is the price of one yard 1 Ans. l,586dol.+ 
 
 NOTE. When decimals or whole numbers, are to be 
 divided by 10, 100, 1000, &c. (viz. unity with ciphers,) 
 it is performed by removing the separatrix in the divi- 
 dend, so many places towards the left hand, as there are 
 ciphers in the divisor. 
 
 EXAMPLES. 
 
 ( 10 the quotient is 74,8dols. 
 '< 
 
 $748 divided by I 100 - - 7,48dols. 
 ( 1000 - ,748dol. 
 
 REDUCTION OF DECIMALS. 
 
 CASE 1. 
 
 To reduce a Vulgar Fraction to its equivalent decimal. 
 
 RULE. Divide the numerator by the denominator, an- 
 nexing as many ciphers as are necessary ; and the quo- 
 tient will be the decimal required. 
 
 EXAMPLES. 
 
 1. Reduce A- to a decimal. 
 
 24)5,0000(,20833+Ans. 
 
 48 
 
 200 
 192 
 
 80 
 
 72 
 
 80 
 
 72 
 
 8 Remainder. 
 
 2. Required the equivalent decimal expressions for | 
 and f . Ans. ,25 ,5 and ,75. 
 
 3. Reduce f to a decimal. Ans. ,375. 
 
 4. Reduce ^ and ff to decimals. Ans. ,04 and ,407.+ 
 
DECIMAL FRACTIONS. 85 
 
 5. Reduce f arid T | T to decimals. 
 
 Ans. ,88 and ,00689.+ 
 
 CASE II. 
 
 To reduce numbers of different denominations to their equiv- 
 alent decimal values. 
 
 RULE 1. Write the given numbers perpendicularly 
 under each other for dividends, proceeding orderly from 
 the least to the greatest. 
 
 2. Opposite to each dividend, on the left hand, place 
 such a number for a divisor, as will bring it to the next 
 superiour denomination, and draw a line between them. 
 
 3. Begin with the uppermost, and write the quotient 
 of each division, as decimal parts, on the right hand of 
 the dividend next below it ; and so on until they are all 
 used, and the last quotient will be the decimal sought. 
 
 EXAMPLES. 
 
 1. Reduce 15s. 9fd. to the decimal of a pound. 
 3, 
 
 20 
 
 9,75 
 15,8125 
 
 ,790625 the decimal required. 
 
 2. Reduce 19s. to the decimal of a pound. 
 
 Ans. ,95. 
 
 3. Reduce 10s. 9d. Iqr. to the decimal of a pound. 
 
 Ans. ,5385416.+ 
 
 4. Reduce Id. 2qrs. to the decimal of a shilling. 
 
 Ans. ,125. 
 
 5. Reduce lOoz. 18dwt. 16grs. to the decimal of a Jk 
 Troy. Ans. ,911111.+ 
 
 6. Reduce lOoz. 14drs. to the decimal of a hundred 
 weight. Ans. ,0060686.+ 
 
 7. Reduce 3 rods 2^-feet, 6 inches to the decimal of a 
 mile. Ans. ,00994318.+ 
 
 8. Reduce 1 pint to the decimal of a gallon. 
 
 Ans. ,125. 
 
 9. Reduce 2 months, 2 weeks, 2 days to the decimal 
 of a year. Ans. ,197302.+ 
 
 H 
 
86 DECIMAL FRACTIONS, 
 
 10. Reduce 3s. 4d. New-England currency to the de- 
 cimal of a dollar. Ans. ,555555.+ 
 
 CASE III. 
 To reduce any number of shillings, pence and farthings 
 
 by inspection to the decimal of a pound. 
 RULE. Write half the greatest even number of shil- 
 lings for the first decimal figure, and let the farthings in 
 the given pence and farthings possess the second and 
 third places ; observing to increase the second place by 
 5, if the shillings be odd ; and the third place by one, 
 when the farthings exceed 12, and by 2, when they ex- 
 ceed 36. 
 
 EXAMPLES. 
 
 1. Find the decimal of 15s. Sd. by inspection. 
 
 ,7=i of 14s. 
 ,05 for the odd shilling. 
 ,034=farthings in 8d. 
 ,001 for excess above 12. 
 
 ,785=decimal required. 
 
 2. Find by inspection the decimal of 12s. 6d. 
 
 Ans. ,628. 
 
 3. Find by inspection the decimal of 18s. lO^d. 
 
 Ans. ,943. 
 
 4. Find by inspection, and add together the decimal of 
 13s. 6d., 9s., Is. 9d., 5d. f, and l^d. Ans. .1,242.+ 
 
 CASE IV. 
 Tojind the value of any given decimal in terms of the 
 
 integer. 
 
 RULE 1. Multiply the decimal by the number of parts 
 in the next less denomination, and cut off as many places 
 for a remainder on the right hand, as there are places in 
 the given decimal. 
 
 2. Multiply the remainder by the parts in the next in- 
 feriour denomination, and cut off for a remainder as be- 
 fore. 
 
 3. Proceed in this manner through all the parts of the 
 integer, and the several denominations, standing on the 
 left hand, will make the answer. 
 
DECIMAL FRACTIONS. 87 
 
 EXAMPLES. 
 
 1. What is the value of ,74*26 of a pound? 
 ,7426 
 20 
 
 s. 14,8520 
 
 12 
 
 d. 10,2240 
 4 
 
 ,8960 Ans. 14s. 10|d.+ 
 
 2. What is the value of ,384 of a shilling? Ans. 4.+ 
 
 3. What is the value of ,6725cwt., at 251b. a qr. 
 
 Ans. 2qrs. 171b. 4oz. 
 
 4. What is the value of ,61 of a tun of wine ? 
 
 Ans. 2hhds. 27gals. 2qts. lpt.+ 
 
 5. What is the value of ,25 of an hour ? 
 
 Ans. 15 minutes. 
 
 6. What is the value of ,857 of a day ? 
 
 Ans. 20h. 34m. 4s. + 
 
 7. What is the value of ,125 of a gallon ? 
 
 Ans. 1 pint. 
 
 CASE V. . 
 To find the value of any decimal of a pound by inspection. 
 
 RULE. Double the first figure or place of lOths for 
 shillings, and if the second be five, or more than 5, add 
 another shilling, then call the figures in the second and 
 third places, after the 5 (if contained) is deducted, far- 
 things ; abating 1 if their number is more than 12, and 
 two if more than 36 ; the result will be the answer. 
 
 EXAMPLES. 
 
 1. Find the value of ,785^ by inspection. 
 I4s.=double 7. 
 Is. for 5 in the place of lOths. 
 8f =35 farthings. 
 i abated for the excess above 12. 
 
 15s. 8d. Answer. 
 
88 REDUCTION OF CURRENCIES. 
 
 2. Find the value of ,976 by inspection. 
 
 Ans. 19s. 
 
 3. Find the value of ,542< by inspection. 
 
 Ans. 10s. lOd. 
 
 4. Find by inspection and add together the values of 
 
 , ,875^, ,096, ,763^, and ,008. 
 
 Ans. 1 19s. 
 
 REDUCTION OF CURRENCIES. 
 
 CASE I. 
 
 To reduce the currencies of the several United States, where 
 a dollar is an even number of shillings, to Federal Money. 
 
 RULE 1. When the sum consists of pounds only, an- 
 nex a cipher to it, and divide by half the number of shil- 
 lings in a dollar ; and the quotient will be dollars ; if 
 there be a remainder, annex ciphers to it, and divide 
 again, by which you will get the cents, &c. 
 
 2. But if the sum consists of pounds, shillings, pence, 
 &c. bring it into shillings, and reduce the pence and far- 
 things to a decimal of a shilling ; annex said decimal to 
 the shillings, with a decimal point between ; then divide 
 the whole by the number of shillings contained in a dol- 
 lar, and the quotient will be dollars, cents, mills, &c. 
 
 EXAMPLES. 
 
 1. Reduce ,73 New-England, Virginia, &c. currency, 
 to Federal Money. 
 
 3)730 
 
 $243,33cts. Ans. 
 
 2. Reduce .45 15s. 7^d. New-England currency to 
 Federal Money. 
 
 20 
 
 s. -- d. d.=,5 
 
 A dollar =6)915, 625 12)7,500 
 
 - , - . - d. 
 
 $152, 604+ Ans. ,625 decimal =7 
 c. m. 
 
REDUCTION OF CURRENCIES. 89 
 
 3. Reduce ^105 14s. 3fd. New-York and North-Car- ' 
 olina currency to Federal Money. 
 
 . s. d. 
 
 105 14 3f f =,75 of a penny. 
 
 20 d. 
 
 s. 12)3,7500 
 
 Adollar=8)2114, 3125 
 
 ,3125 decimal =f 
 
 $264, 2S9 T o 6 <y+ Ans. 
 c. m. 
 
 4. Reduce <431 New-York currency to Federal 
 Money. 
 
 4)4310 
 
 $1077, 50cft>. Ans. 
 
 RULE 2. Bring the given sum into a decimal expres- 
 sion by inspection, as in case 3 of decimal fractions ; 
 then divide the whole by ,3 in New-England, &c. cur- 
 rency, and by ,4 in New-York, &c. currency ; and the 
 quotient will be dollars, cents, &c. 
 
 EXAMPLES. 
 
 1. Reduce <54 8s. 5d. New-England currency to 
 Federal Money. . 
 
 ,3)54,415 decimally expressed. 
 
 $18l,3Sc.+ Ans. 
 
 2. Reduce 75. life?. New-England currency to Fed- 
 eral Money. 
 
 7s. llr/.=<0,399 decimally expressed. 
 Then ,3),399 
 
 $l,33c. Ans. 
 
 3. Reduce <513 16s. lOt/. New-York, &c. currency 
 to Federal Money. 
 
 . 
 ,4)513,842 decimally expressed. 
 
 $J284,60c. Ans. 
 H3 
 
90 REDUCTION OF CURRENCIES. 
 
 4. Reduce 19s. 5d. New-York, &c. currency to Fed- 
 eral Money. <. 
 
 ,4)0,974 decimal of 19s. 5d. 
 
 $2, 43c. Ans. 
 
 NOTE. By the preceding rule, you may carry on the 
 decimal to any degree of exactness ; but in ordinary prac- 
 tice, the following Contraction may perhaps be useful. 
 
 RULE 3. To the shillings contained in the given sum, 
 annex 8 times the given pence, increasing the product by 
 2 ; then divide the whole by the number of shillings con- 
 tained in a dollar, and the quotient will be cents. 
 
 EXAMPLES. 
 
 1. Reduce 45s. 6d. New-England, &c. currency, to 
 Federal Money. 
 
 s. 6x8+2=50 to be annexed. 
 
 6)45,50 
 
 $7,58f c. Ans. or 6)4550 
 
 $c. 
 
 758+ cents =7,58 
 
 2. Reduce 2 10s. 9d. New-York, &c. currency, to 
 Federal Money. 2 I0s.=50s, 
 
 9x8+2=74 to be annexed. 
 Then 8)5074 s. 
 
 $ c. Or thus 8)50,74 
 
 634+cents=6,34 
 
 $6,34|c. Ans. 
 
 NOTE. When there are no pence given in the sum, 
 you must annex two ciphers to the shillings ; then divide 
 as before, &c. 
 
 3. Reduce <3 5s. New-England currency, to Federal 
 Money. 
 
 35s.=65s. Then 6)6500 
 
 $ c. 
 
 1083cts.= 10,83 Ans. 
 
REDUCTION OF CURRENCIES. 91 
 
 CASE? II. 
 
 To reduce the currency of New- Jersey, Pennsylvania, 
 Delaware, and Maryland, to Federal Money. 
 
 Rule. Multiply the given sum by 8, and divide the 
 product by 3, and the quotient will be dollars, &c. 
 
 EXAMPLES. 
 
 1. Reduce <245 New-Jersey, &c. currency, to Fed- 
 eral Money. 
 
 f 653,33c. Ans, 
 
 NOTE. When there are shillings, pence, &c. in the 
 given sum, reduce them to the decimal of a pound, then 
 multiply and divide as in the preceding question. 
 
 2. Reduce =36 Us. 8^d. New-Jersey, &c. currency, 
 to Federal Money. 
 
 j36,5354 decimal value. 
 
 8 
 
 3)292,6832 
 
 $97,561-^+ Ans. 
 c. m. 
 
 3. Reduce 17s, 9d. New-Jersey, &c. to Federal Mon- 
 ey. Ans. $2,36c. 6f m. 
 
 CASE III. 
 
 To reduce the currency of South- Carolina, and Georgia, 
 to Federal Money. 
 
 RULE. Multiply the given sum by 30, and divide the 
 product by 7 ; the quotient will be dollars, cents, &c. 
 Or, multiply by 3 and divide by ,7. 
 
 EXAMPLES. 
 
 1. Reduce <100 South-Carolina and Georgia, curren- 
 cy, to Federal Money. 
 
 ^100x30=3000; 3000--7 =$428,57 14+ Ans. 
 
92 REDUCTION OF CURRENCIES. 
 
 2. Reduce 5 J 6s. 9f Georgia currency to Feder- 
 al Money. 
 
 54,8406 decimal expression, 
 30 
 
 7)1645,2180 
 
 $235,0311+ Ans. 
 
 3. Reduce Us. 6d. South-Carolina, &c. to Federal 
 Money. Ans. $2 46c. 4m. + 
 
 CASE IV. 
 
 To reduce the currency of Canada and Nova-Scotia, to 
 Federal Money. 
 
 RULE. Multiply the given sum by 4, if it contain 
 pounds only, and the product will be dollars. If it con- 
 tain shillings, reduce the whole to shillings, and divide by 
 5 ; if it contain pence, reduce the whole to pence, and 
 divide by 60 ; and the quotient, in either case, will be dol- 
 lars ; to the remainders, if there be any, annex ciphers, 
 and continue the division, by which you will obtain the 
 cents and mills. 
 
 Or, when the given sum contains shillings, pence, &c. 
 reduce them to the decimal of a pound, annex the deci- 
 mal to the pounds, and multiply the whole by 4 ; the pro- 
 duct will be dollars, cents, &c. 
 
 EXAMPLES. 
 
 1. Reduce ^6125 Canada and Nova-Scotia currency, to 
 Federal Money. 
 
 125 
 4 
 
 $500 Ans. 
 
 2. Reduce ^68 14s. Nova-Scotia currency, to Federal 
 Money. 
 
 . s. . , 
 
 68 14 Or 68, 7 decimal expression, 
 
 20 4 
 
 5) 1 374 $274,8 dimes. Ans. 
 
 f 274,80cts. Ans. 
 
DEDUCTION OF CURRENCIES. 93 
 
 3. Reduce 45 17s. 9d. Canada and Nova-Scotia cur- 
 rency to Federal Money. 
 
 . s. d. 
 
 45 179 . 
 
 20 Or 45,8875 decimal expression. 
 
 4 
 
 917 
 
 12 $183,5500 
 
 6,0)1101,3 
 
 $183,55cts. Ans. 
 
 4. Reduce 58 13s. 6d. Canada, &c. to Federal 
 Money. 
 
 . s. d. 
 
 58 13 6 =,5 of a penny. 
 20 . 
 
 Or 58,67708 decimal expression. 
 
 1173 4 
 
 12 $. c.m. 
 
 8234,70832=234,708 T 3 <&+Ans. 
 6,0)1403,2^ 
 
 $234,708 Ans. 
 c. m. 
 
 5. Reduce .528 17s. 8d. Canada, &c. to Federal 
 Money. Ans. $2115,53cts.+ 
 
 6. Reduce 1 2s. 6d. Nova-Scotia, &c. money, to 
 Federal Money. Ans. $4,50cts. 
 
 7. Reduce 13s. ll^d. Nova-Scotia, &c. money, to 
 Federal Money. Ans. $2,79cts.+ 
 
 CASE V. 
 
 To reduce the money of Great Britain to Federal Money. 
 
 RULE. If the given sum be pounds only, multiply by 
 40, and divide by 9, or multiply by 4, and divide by ,9 ; 
 the quotient will be dollars ; if there be any remainder, 
 annex ciphers to it, and continue the division ; the quo- 
 tient will be cents, &c. But if it consist of pounds and 
 shillings, reduce it to shillings, then double them, and 
 divide as before. And if it contain pounds, shillings, 
 
94 REDUCTION OF CURRENCIES. 
 
 and pence, reduce it to pence, and divide by 54, the num- 
 ber of pence in a dollar. 
 
 Or, when the sum consists of pounds, shillings, and 
 pence, reduce the shillings, &/c.to the decimal of a pound, 
 then multiply the whole by '40, and divide by 9. 
 
 EXAMPLES. 
 
 1. Reduce 36 sterling into Federal Money. 
 
 36x40 
 
 =$160 Ans. 
 
 9 
 
 2. Reduce ^636 9s. sterling into Federal Money. 
 
 . s. . 
 
 36 9 Or 36,45 decimal expression. 
 
 20 40 
 
 729 9)1458,00 
 
 o 
 
 $162 Ans. 
 
 9)1458 doubled. 
 
 $162 Ans. 
 3. Reduce ,579 17s. 9d. sterling into Federal Money. 
 
 . s. d. 
 579 17 9 . 
 
 20 Or 579,8875 decimal expression, 
 40 
 
 11597 
 
 12 9)23195,5000 
 
 c. m. 
 
 54)139173(2577,277+ An.$2577,2777+Ans. 
 
 4. Reduce <100 sterling money, to Federal Money. 
 
 Ans. $444,44|c. 
 
 To reduce Federal Money to the currency of 
 
 Neu^England, j Multiply the given sum by ,3 
 Virginia, I J J & the pro duct will be pounds, 
 
 Kentucky, and f* \ ^ jg^to of a pound. 
 
 Tennessee. ) 
 
 7 , . . ( Multiply the given sum by ,4 
 "*?%* F A H | and the product will be pounds, 
 North-Carohna. 5 \ and dec imals of a pound. 
 
REDUCTION OF CURRENCIES. 
 
 96 
 
 3. 
 
 4. 
 
 New-Jersey, 
 Pennsylvania, 
 Delaware and 
 
 Maryland. 
 
 South-Carolina ) j> 
 and Georgia. ] & 
 
 Multiply the given sum by 3, 
 and divide the product by 8, 
 and the quotient will be pounds 
 and decimals of a pound. 
 
 Multiply the given sum by ,7 
 and divide the product by 3, 
 and the quotient will be pounds 
 and decimals of a pound. 
 
 EXAMPLES IJV THE FOREGOING RULES. 
 
 1. Reduce $152,6Qcts. to New-England currency. 
 152,00 
 ,3 
 
 .45,780 Ans. ,45 15s. 7d. T V 
 
 20 But the value of any decimal of a pound, 
 
 may be found by inspection, as in case 5, 
 
 s. 15,600 of Decimal Fractions, page 87. 
 12 
 
 d. 7,200 
 
 2. Reduce $198 into Virginia, &c. currency. 
 196 
 ,3 
 
 58,8 Ans. =^58 16s. 
 
 3. In $629, how many pounds New- York, &c. currency ? 
 629 
 ,4 
 
 .=^251 12s. 
 
 4. Bring $I10,51cts. into New-Jersey, &c. currency. 
 110,51 
 
 3 Double ,4 -8s. Then take 2 from 41 = 
 - 39 for fart lungs =9d. 3qrs. See Case 5 
 8)331,53 of Decimals, page 87. 
 
 <41,441 Ans. =41 8s. 9fd. by inspection. 
 
96 KEDUCTION OF CURRENCIES. 
 
 5. How many pounds, &c. South-Carolina, &c. cur- 
 rency in $65,36cts. Smills ? 
 65,368 
 
 ,7 
 
 3)45,7576 
 
 ^15,25253 Ans.=,15 5s. 
 
 CASE II. 
 
 To reduce Federal Money to Canada and Nova-Scotia 
 currency. 
 
 RULE. Divide the dollars by 4, and the quotient will 
 be pounds ; to the remainder, if there be any, annex the 
 cents, &c. and to that number, annex a cipher ; then halve 
 that number, and cut off the left hand figure or figures 
 less than 20, for shillings ; the remaining figure or figures 
 multiply by 12, and cut off just as many right hand fig- 
 ures from the product as you multiply ; the left hand 
 ones are pence, &c. 
 
 Or, multiply the given sum by 60, the number of pence 
 in a dollar, and if it contains cents, cut off two figures on 
 the right, if mills, three ; those on the left are pence, 
 which must be reduced into pounds. The figures cut off 
 will be decimals of a penny. 
 
 Or, lastly, divide the given sum by 4, and the quotient 
 will be pounds, and decimals of a pound. 
 
 EXAMPLES. 
 
 1. Reduce $183,55cts. into Canada and Nova-Scotia 
 Money. 
 
 $ 
 4)183 
 
 ,45 and 3 remain to be placed before the 
 55 cents. 355, to which annex a cipher. 3550 halved, 
 or divided by 2=1775, of which cut off the two left hand 
 figures, as they are less than 20, which are shillings 17,75. 
 75x12=900, of which cut off the two right hand figures 
 because you multiply two =9,00; and the 9 on the left 
 are pence ; the answer is, therefore, ^45 17s. 9d. 
 
REDUCTION OF CURRENCIES. J)7 
 
 By the second method : 183,55x60=1101300, from 
 which cut off the two right hand figures, and it is 11013, 
 which are pence. These reduced, are 45 17s. 9d. Ans. 
 
 By the last method : $. c. 
 4)183,55 
 
 45,8875 =45 17s. 9d. Ans. 
 
 2. Bring S741 into Nova-Scotia currency. 
 
 Aris. 185 5s. 
 
 3. What sum, Nova-Scotia money, is equal to $311, 
 75cts. ? Ans. ,77 18s. 9d. 
 
 4. In $2907,56cts., how much Nova-Scotia money 1 
 
 Ans. 726 l?s. 9d.+ 
 
 5. How many pounds, &c. Canada money are in 
 $2114,50cts. ? Ans. ,528 12s. 6d. 
 
 CASE III. 
 
 To reduce Federal Money to the money of Great Britain. 
 
 RULE. Multiply the given sum by 9, and divide the 
 product by 40 ; or multiply the given sum by ,9 and di- 
 vide the product by 4 ; and, in either case, the quotient 
 will be the answer in pounds, and decimals of a pound. 
 
 EXAMPLES. 
 
 1. Reduce $l83,55cts. into sterling money. 
 183,55 
 9 
 
 Or 183,55 
 
 4,0)165,195 ,9 
 
 41,29875 4)165,195 
 20 
 
 41,29875=^41 5s. lld. by in- 
 
 s.5,97500 spection, 
 
 12 
 
 d. 11,700 
 4 
 
 s. lid. 2,8qrs. Ans. 
 
 2. Bring $247,44c. 5m. into English money. 
 
 Ans. 55 13s. 
 
 3, Show the value of $1000 in British money. 
 
 Ans. 225. 
 
98 RULE OF THREE. 
 
 4. Tell me what sum, in sterling money, is just equal 
 to $2466,33cts. 3mills. Ans. ^554 18s. 6d. 
 
 RULE OF THREE. 
 
 THE RULE OF THREE teaches to find a number having 
 the same proportion to a given number, that two other 
 given numbers have between themselves. For this reason 
 it is sometimes called the Rule of Proportion. It is call- 
 ed the Rule of Three, because in each of its questions 
 there are given three numbers at least. And because of 
 its excellent and extensive use, it has been often named 
 the Golden Rule. 
 
 RULE. Write down the number, which is of the same 
 kind with the answer or number required. 
 
 Consider whether the answer ought to be greater or less 
 than this number ; if greater, write the greater of the 
 two remaining numbers on the right hand of it for the 
 third, and the other on the left for the first number or 
 term ; but if it ought to be less, write the less of the two 
 remaining numbers in the third place, and the other in 
 the first. 
 
 Multiply the second and third terms together, divide the 
 product by the first, and the quotient will be the answer. 
 
 NOTE 1. It is sometimes most convenient to multiply 
 and divide as in Compound Multiplication and Division. 
 But when it is not, then reduce each of the compound 
 terms to the lowest denomination mentioned in it, and 
 reduce the first and third to the same denomination ; then 
 will the answer be of the same denomination with the 
 second term. And the answer may afterwards be brought 
 to any denomination required. 
 
 2. When there happens to be a remainder after the 
 division, reduce it to the name next below the last quo- 
 tient, and divide by the same divisor ; so shall the quo- 
 tient be so many of the said next denomination ; do this 
 as long as there is any remainder, or till you have reduc- 
 ed it to the least name, and all the quotients together 
 will be the answer. 
 
RULE OF THREE. 99 
 
 3. If the first term, arid either the 2d or 3d, can be di- 
 vided by any number, without remainder, let them be di- 
 vided, and the quotients used instead of them. 
 
 4. There are four other methods of operation besides 
 the general one above delivered, any of which, when 
 possible, performs the work much shorter than it. They 
 are thus : 
 
 First, Divide the 2d term by the 1st, multiply the quo- 
 tient by the 3d, and the product will be the answer. 
 
 Second, Divide the 3d term by the 1st, multiply the 
 quotient by the 2d, and the product will be the answer. 
 
 Third, Divide the 1st term by the 2d, divide the 3d by 
 the quotient, and the last quotient will be the answer. 
 
 Fourth, Divide the 1st term by the 3d, divide the 2d 
 by the quotient, and the last quotient will be the answer. 
 
 Two or more statings are sometimes necessary, which 
 may always be known from the nature of the question. 
 
 The method of proof is by inverting the question. 
 
 EXAMPLES. 
 
 1. What is the value of Zjfo. 6oz. I9dwt. of gold, at 
 ^3 19s. lid. an ounce? 
 
 oz. . s. d. }>. oz. dwt. 
 
 1 : 3 19 11 : : 2 6 19 
 20 20 12 
 
 20 79 30 
 
 12 20 
 
 959 619 
 
 619 
 
 8631 
 959 
 5754 
 
 2,0)59362,1 
 
 12)2968 1 ^ Pence, which divided by 12 and 
 20, gives the answer in pounds, 
 2,0)247,3s. 5d. &c. 
 
 Ans. 123 13s. 
 
100 KULE OF THREE. 
 
 2. If 9}fc. of tobacco cost I dollar 20 cents, what will 
 . cost ? 
 
 ft- 8- cts. ft. 
 
 9 : 1,20 : : 25 
 120 
 
 9)3000 
 
 3,33 Ans. 
 
 3. If 25ft. of tobacco cost $3,33, what will Oft. cost ? 
 ft- * cts. ft. 
 
 25 : 3,33i : : 9 
 9 
 
 25)3000(120 cents, or $1,20 the An. 
 25 
 
 50 
 50 
 
 
 
 4. What is the value of a firkin of butter containing 
 56ft. at I0d. per pound ? 
 
 ft. 
 : 56 
 
 <2 9 the Answer. 
 
 5. If 7cwt. Iqr. of sugar cost 36dols. 10c., what will 
 be the price of 43cwt. 2qr, ? Ans. g216, 60c. 
 
 cwt.gr. 
 : 43 2 
 4 
 
 174 
 
 29)628140(21660cts.=$216, 60cts. 
 
 cwt. qr. 
 
 $. cts. 
 
 7 I 
 
 : 36,10 
 
 4 
 
 17 4 
 
 29 
 
 1444 
 
 
 2527 
 
 
 361 
 
RULE OF THREE. 101 
 
 6. If 6 horses eat 21 bushels of oats in a month, haw 
 many bushels will 20 horses eaf in the same time.? 
 
 Hor. Bus. Hor. Bus. 
 
 6 : 21 :: 20 : 70 Ans. 
 
 7. A man bought sheep Idol, llcts per head, to the 
 amount of 51dol. 6cts : how many sheep did he buy 1 
 
 $ cts. sh. $ cts. sh, 
 
 1,11 : 1 :: 51,06 : 46 Ans. 
 
 8. What is the value of an cwt. of sugar at 5^d. per 
 ft., 25ft. a qr. ? 
 
 ft- d. qr. ft. &s.d- 
 
 1 : 52 : : 100 : 2 5 10 Ans. 
 
 9. How much in length of that which is 4J inches 
 broad will make a square foot ? 
 
 Breadth. Length. Breadth. Length. 
 4 : 12 :': 12 : 2ft. Sin. Ans. 
 
 10. Bought 6 casks of raisins, each weighing Icwt. lq. 
 12^1b. ; what will they come to at 2 Is. fed. per cwt. ? 
 
 Cwt. s. d. Cwt. qr. fb- s. d. 
 
 1: 2 18 :: 1 I 12^ + 6 : 17 4+ Ans. 
 
 11. If a man spend 2 dollars 45 cents a week, what 
 will it amount to in a year ? 
 
 d'ys. $. cts. d'ys. $. cts. 
 
 7 : 2,45 : : 365 : 127,75 Ans. 
 
 12. What is the value of a pipe of wine at 10-J-d-. per 
 pint ? 
 
 Pint. d. Pipe. . s. 
 
 1 : 10 :: 1 : 44 2 Ans. 
 
 13. How many quarters of corn can I buy for 280 dol- 
 lars, at f of a dollar per bushel ? 
 
 Ans. 52 quarters, 4 bushels. 
 
 14. What is the value of 2qrs. Ina. velvet, at 19s. 8^d. 
 per ell English ? Ans. 8s. lO^d. T V 
 
 15. Suppose 18 yards of broadcloth l^yds. wide is to 
 be lined with shalloon that is f of a yard wide ; how many 
 yards of shalloon will be sufficient ! Ans. 36 yds. 
 
 16. If 52 yards of cloth cost 156 dollars, how much 
 will 4 yards cost ? Ans. 12 dollars. 
 
 17. Bought 36 yards of cloth for 108 dollars, and sold 
 the same at 3| dollars per yard ; how much did I gain ? 
 
 Ans. 18 dollars. 
 I 2 
 
ULE OF THREE. 
 
 18. If 7yds. of ribbon cost 3s. 4d.', what will 126yds. 
 cost ? Ans. 3. 
 
 19. If a man earn 64*dollars in 4 months, how long 
 must he work at the sanie rate to pay a debt of 300 dol- 
 lars ? Ans. 18 months, 3 weeks. 
 
 20. If an ounce of silver be worth 1 dollar, 10 cents, 
 what is the value of 10 silver spoons, each weighing loz. 
 4 pennyweights ? Ans. 13dolls. 20cts. 
 
 21. If 8f yards cost 4 dollars 20 cents, what will 13 
 yards cost I Ans. Gdolls, 48cts. 
 
 22. How long will it take 5 men to do the same work 
 which 37 men can do in 15 days 1 Ans. Ill days. 
 
 23. what will 4 hogsheads of wine come to contain- 
 ing, viz. 79|, 84^, 101J, and 112 gallons, at 6s. 9d. per 
 gallon ? Ans. 127 4s. 9d. 
 
 24. Bought 3hhds. of sugar, each weighing 8cwt. Iqr. 
 12fty at 7dolls. 26cts. per cwt ; what come they to, 25fc. 
 to the qr ? Ans. $l82,29c. 8f m. 
 
 25. If a chest of hyson tea weighing 79{fe. neat, cost 
 32 lls. 9d., what is it per pound 1 Ans. 8s. 3d. 
 
 26. B owes 2119 17s. 6d., and he is worth but ,1324 
 18s. 5d. ; if he delivers this to his creditors, how much 
 do they receive on a pound ? Ans. 12s. 6d. 
 
 % 27. A merchant failing in trade, owes in all 29475 dol- 
 lars, and delivers up his whole property worth 21894 dol- 
 lars, 3 cents ; how much per cent, does he pay ; and what 
 is B's loss to whom he owed 325 dollars ? 
 
 Ans. He pays $74,28cts. per cent, 
 and B loses f 83,59cts. 
 
 28. If a staff, 4 feet, 8 inches in length, cast a shadow 
 6 feet ; how high is that steeple whose shadow is 153 feet ? 
 
 Ans. 119 feet. 
 
 29. Bought 270 quintals codfish for 780 dollars ; freight 
 37 dollars, 70cts. ; wharfage, truckage and other expenses 
 30 dollars, 60cts. ; at what must I sell them per quintal, 
 so as to gain 143 dollars on the whole ? 
 
 Ans. 3,67cts. 1m. + 
 
 30 If f of a farm cost $1081, what is the whole 
 worth 1 
 
 fif. $. fif. $. cts. m. 
 
 3 : 1081 : : 5 : 1801,66 6+Ans. 
 
RULE OF THREE- 103 
 
 31. If a man spend 46 cents a day, what will it amount 
 to in a year ? Ans. $ 1 67,90. 
 
 32. Lent a friend 292 dollars for six months ; sometime 
 afterward he lent me 403 dollars ; how long must I keep 
 it to balance the favour ? Ans. 4m. lw^2d.+ 
 
 33. If 100 dollars gain 6 dollars interest in one year, 
 how much will 480 dollars gain in the same time ? 
 
 Ans. 828,80. 
 
 34. If 480 dollars, gain 28 dollars, 80 cents in one year, 
 > how much will it gain in 87 days ? 
 
 Ans. $6,86cts. 4m. + 
 
 35. How much land, at, 2 dollars, 50 cents per acre, 
 must be given in exchange for 360 acres, at 3 dollars, 75 
 cents per acre ? Ans. 540 acres. 
 
 36. Bought a silver cup weighing 9 ounces 4 penny- 
 weights 16 grains, for 3 2s. 3d. 3qrs.f ; what was it per 
 ounce ? Ans. 6s. 9d. 
 
 37. There is a cistern which has four cocks ; the first 
 will empty it in 10 minutes, the second in 20 minutes, 
 the third in 40 minutes, the fourth in 80 minutes; in 
 what time will all four running together empty it ? 
 
 Ans. 5min. 20sec. 
 
 38. A hired two men,,B and C, to cut wood for 50cts. 
 per cord ; B could cut a cord in 4 hours, C in 6 hours ; 
 how long would it take both to cut 1 cord ? 
 
 Ans. 2 hours, 24 minutes. 
 
 39. If, when wheat is 6s. 3d. per bushel, the penny 
 loaf weigh 9 ounces, what ought it to weigh when wheat 
 is 8s. 2d. per bushel ? Ans. 6oz. 13drs.+ 
 
 40. When a man's yearly income is 949 dollars, how 
 much is it per day ? Ans. $2,60cts. 
 
 41. What is the commission on 1525 dollars at 4|dolls. 
 per cent. ? Ans. g6S,62cts. 5m. 
 
 42. What will 374 feet of boards come to at l^cerits 
 per foot! Ans. $5,61 cts. 
 
 43. What will 39 thousand 6 hundred and 30 casts of 
 staves come to at 15 dollars 50 cents per thousand ? 
 
 Ans. $6l4,73cts. 
 
 NOTE. 2 staves make 1 cast; 50 casts 1 hundred; 
 10 hundred 1 thousand ; in Maine, by a late law of this 
 State. 
 
104 RULE OF THREE. 
 
 44. If the inventory of a town be 358400 dollars, upon 
 which there is assessed a tax of 850 dollars, what will it 
 be on a dollar ; and what will B's tax be, whose estate 
 in that town is valued at 1792 dollars ? 
 
 ( 2mills T 3 o^+ on a dollar, and 
 s \ B's tax will be g4,25cts. 
 
 45. What will the charter of a ship of 306 tons amount 
 to, from May 28 to October 10th following, at 2 dollars 
 per ton, per month of 30 days 1 . 
 
 Ans. 2774 dollars, 40 cents. 
 
 NOTE. The days of receiving and discharging are 
 both included. 
 
 46. If 4^ hundred weight may be carried 36 miles for 
 35s. how many pounds can I have carried 20 miles for the 
 same money, 25}fo. a quarter. Ans. 810ffc. 
 
 47. Sold a ship for 537, and I owned f of her ; what 
 was my part of the money 1 Ans. <201 7s. 6d. 
 
 48. What quantity of water must I add to a pipe of 
 mountain wine valued at ,33, to reduce the first cost to 
 4s. 63. per gallon 1 Ans. 20f gallons. 
 
 49. A and B depart from the same place, and travel 
 the same road ; but A goes 6 days before B, at the rate of 
 21 miles a day ; B follows at the rate of 28 miles a day ; 
 in what time and distance will he overtake A 1 
 
 . MS days. 
 3 * \ 504 miles. 
 
 50. A factor bought a certain quantity of broadcloth 
 and drugget, which, together, cost 81 ; the quantity of 
 broadcloth was 50 yards, at 18s. per yard, and for every 
 5 yards of broadcloth he had 9 yards of drugget ; I de- 
 mand how many yards of drugget he had, arid what it 
 cost him per yard ! Ans. 90yds. at 8s. per yd. 
 
 51. If 60 gallons of water, in one hour, fall into a cis- 
 tern containing 300 gallons, and by a pipe in the cistern, 
 35 gallons run out in forty minutes ; in what time will it 
 be filled 1 Ans. 40 hours. 
 
 52. How many yards of .cloth 3qrs. wide, will be equal 
 in measure to 30 yards, 5qrs. wide. Ans. 50yds. 
 
 53. Bought a pipe of wine for 84 dollars, and found it 
 
PRACTICE. 105 
 
 leaked out 17 gallons; I sold the remainder at 12cts. a 
 pint ; did I gain or lose, and how much ? 
 
 Ans. I gained 30 dols. 
 
 54. How many yards of paper, 3 quarters wide, will 
 paper a room, 30 feet long, 24 wide, and 12 high, deduct- 
 ing 81 square feet for fire-place, door, and windows 1 
 
 Ans. 180 yards. 
 
 55. A garrison consisting of 1500 men, being besieg- 
 ed, have provisions for three months only ; but it being 
 necessary they should hold out five months, how many 
 men must depart, that the same provisions may serve that 
 time ? Ans, 600 men. 
 
 56. A regiment of soldiers consisting of 1000 men, are 
 to have new coats, and each coat is to contain 2yds. and 
 Iqr. of cloth that is 5 quarters wide ; how much shalloon, 
 that is 3 quarters wide, will line them 1 Ans. 3750yds. 
 
 57. A merchant shipped for the West-Indies 39000 
 feet of boards, at $8,20cts. per M. ; 300 quintals offish, 
 at $2,60cts. per quint. ; 15000 shingles, at $2,20cts. per 
 M. ; 34000 hoops, at gl,60cts per M. ; and glOOO in 
 cash; and in return, he had 3000 ffo. of indigo at 56 
 cents per jfo. 2580 gallons of molasses, at 20cts. per 
 gal. ; 1000 pounds of coffee, at 18cts. per jfc. ; and 18cwt. 
 of sugar, at $4,50cts. per cwt. ; and his charges on the 
 voyage were $l53,80cts. Did he gain, or lose, by this 
 voyage? Ans. He gained SI 16. 
 
 PRACTICE. 
 
 PRACTICE is a contraction of the Rule of Three, when 
 the first term happens to be a unit, or 1 ; and is a concise 
 method of ascertaining the value of goods, &,c. where 
 money is reckoned in pounds, shillings, and pence ; but, 
 since reckoning in Federal Money, in all kinds of busi- 
 ness, has become universal in our country, it has here 
 grown into almost total disuse. I shall, therefore, pre- 
 sent few examples in this Rule to the attention of the 
 student ; arid these chiefly in Federal Money. A table 
 of aliquot or even parts of weight he will find annexed to 
 Case 2, in Tare and Tret, page 109. 
 
10G 
 
 PRACTICE. 
 
 CASE I. 
 
 When the price is an even part of a pound. 
 RULE. Find the value of the given quantity, at one 
 pound per yard, <fcc. and divide it by that even' part, and 
 the quotient will be the answer in pounds. 
 
 EXAMPLES. 
 
 1. What will 129 yards cost, at 2s. 6d. per yard ? 
 The quantity itself is the price at l per yd. 
 2s. 6d.=of . 
 
 . s. 
 2s. 6d.() 129 10 value at 1 per yd. 
 
 Ans. c16 3s. 9d. value at 2s. 6d. per yd. 
 
 2. What will 461 bushels of oats cost, at 1 s. 8d. per. 
 bushel ? Is. 8d.r= T u of a . Ans. ^38 8s. 4d. 
 
 3. What will 21 1^ bus. of corn cost, at 4s ? 
 
 4s. =|. Ans. M<2 5s. 
 
 4. What will 543 bus. of wheat cost, at 6s. 8d. 7 
 
 6s. 8d.=-J. Ans. ^181. 
 
 5. What will 127 gallons of wine cost, at 3s. 4d. ? 
 
 8s. 4d.=f Ans. .21 3s. 4d. 
 
 6. What will 687^ bushels of salt cost, at 5s. ? 
 
 5s.==J.~ Ans. ^171 17s. 6d. 
 
 CASE II. 
 
 When the price is Federal Money, but the quantity of the 
 
 goods is in several denominations. 
 
 RULE. Multiply the price by the integers in the given 
 quantity, and take parts for the rest from the price of an 
 integer ; which added together, will be the answer. 
 
 ,65cts. per 
 
 1. What cost 9cwt. iqr. 8jfe. of sugar at 
 cwt. $ cts. 
 
 lqr.]i-8, 65 
 9 
 
 lib- 
 
 77, 85 
 2, 1625 
 , 5406+ 
 
 , 0772+ 
 
 $80, 6303+ Ans. =$80,63cts. 
 
PRACTICE. 107 
 
 2. What cost 7cwt. 3qr. 16ft. of raisins, at $9,58cts. 
 per cwt. ? Ans. $75,61cts. 3m. + 
 
 cwt. qr. ft. 
 
 3. 12 7 of Russian iron, at $6,34cts. per. cwt., 
 at 25ft. a qr. ? Ans. $76,52cts. 3m. + 
 
 cwt.qr.lb. 
 
 4. 24 of hemp, at $ll,91cts. per cwt. 1 
 
 Ans. $2,55cts. 2m. + 
 
 5. 1 16 of pig lead, at $6,5 lets, per cwt. at 25ft. 
 a qr. ? Ans. $7,55cts. If m. 
 
 6. 2 21 of sugar, at $lO,24cts., per cwt., at 25ft. 
 a qr. ? Ans. f7,27^cts. 
 
 7. 5 cords, 3ft. Sin. of bark, at $4,50cts. per cord ? 
 
 Ans. $24,56cts. 
 
 8. 3tons, 5cwt. 2qr. of hay, at $25,1 Gets per ton ? 
 
 Ans. $82,20cts. 
 
 9. 1 ton, 9cwt. 3qr. 17ft. of hay, at $4,44cts. 4m. 
 per ton, at 25ft. a qr. ? Ans. $6,64cts. 8m. + 
 
 10. 17jfc. 5oz. I4dwt. of electary, at 83cts. per fc. ? 
 
 Ans. $!4,5Qcts.+ 
 
 11. 32ac. 1 rood, 14 perches of land, at $l,l6cts. per 
 acre? Ans. S37,51cts.+ 
 
 12. I4ac. 3 roods, 5 perches of land, at $10,50cts. per 
 acre 1 Ans. g!55,20cts.+ 
 
 13. 6yds. 3qr. 2na. of cloth, at $7,82cts. 8m. 1 
 
 Ans. $53,8 Jets. 7. 
 
 14. 8gal. 3qts. Ipt. 2gills of wine at $1, 96cts. ? 
 
 Ans. f 17,5lfcts. 
 
 The Rule may be applied to other weights and meas- 
 ures also. But perhaps all such questions may be better 
 wrought by multiplication of decimals. Take, for exam- 
 ple, the 9th question in this Case. What will Iton 9cwt. 
 3qr. 17ft of hay come to, at $4,44cts. 4m. per ton, at 
 25ft. a qr. ? Ton. $ $ c. m. 
 
 Decimal expression 1, 496x4,444 =$6,648224 =6,648-f 
 answer. 
 
108 TARE AND TRET. 
 
 TARE AND TRET. 
 
 TARE AND TRET are practical rules for deducting cer- 
 tain allowances, which are made by merchants and 
 tradesmen in selling their goods by weight. 
 
 Gross weight is the whole weight of any sort of goods, 
 together with the box, barrel, or bag, &c. that contains 
 them. 
 
 Tare is an allowance made to the buyer for the weight 
 of the box, barrel, or bag, &c. which contains the goods 
 bought, and is either at so much per box, &c., at so much 
 per cwt., or at so much in the gross weight. 
 
 Tret is an allowance of 4f^. in every 104ffe. for waste, 
 dust, &c., or ^ of the whole tare=suttle. 
 
 Cloff is an allowance of 2ff). upon every 3cwt. or 336|fc. 
 
 Suttle is the weight when part of the allowance is de- 
 ducted from the gross. 
 
 Neat weight is what remains after -all allowances are 
 made. 
 
 CASE I. 
 
 When the tare is a certain weight per box, barrel, or 
 bag, fyc. 
 
 RULE. Multiply the number of boxes, or barrels, &c. 
 by the tare, and subtract the product from the gross, and 
 the remainder is the neat weight required. 
 
 EXAMPLES. 
 
 1. In 10 casks of alum, each weighing 3cwt. 2qrs. 
 . gross, tare 15ft, per cask, how much neat ? 
 Cwt. qr. ffc. 
 3 2 12 
 10 
 
 36 8 Gross. 
 15xlO=150fb=l 1 10 Tare. 
 
 34 2 26 neat, the answer. 
 
TARE AND TRET. 
 
 109 
 
 2. In 241 barrels of figs, each 3qrs. lOfc. gross, tare 
 10}k. per barrel, how many pounds neat ? 
 
 Ans. 22413ft. 
 
 3. What is the neat weight of 21 hogsheads of tobac- 
 co, each 5cwt. 2qrs. 171b. gross, tare lOOlb. per hogshead, 
 at 251b. to the qr. I Ans. 98cwt. 71b. 
 
 4. What is the neat weight of 4 chests of hyson tea, 
 weighing, gross 961b. 971b. lOllb. and 1031b., tare 201b. 
 per chest I Ans. 3171b. 
 
 CASE II. 
 
 Wlien the tare is a certain weight per cwt. 
 RULE. Divide the gross weight by the aliquot* parts 
 of an cwt. contained in the tare, and subtract the quotient 
 from the gross, and the remainder is the neat weight. 
 
 EXAMPLES. 
 
 1. Gross 372cwt. 3qrs. 171b., tare 161b. per cwt., how 
 much neat ? Cwt. qrs. Ib. 
 
 16Ib. is i)372 3 17 
 
 53 1 2 t^re subtracted. 
 
 319 2 I4flhe answer. 
 
 2. What is the neat weight of 7 barrels of potash, each 
 weighing 4021b. gross, tare lOlb. per cwt., at 251b. to 
 the qr. 1 Ans. 25321b. 9|oz. 
 
 * An aliquot part of any number is such a part of it 
 as, being taken a certain number of times, exactly makes 
 that number. If lOOlb. be taken for a cwt., 501b.=; 
 
 v 
 
 TABLE OF ALIQUOT PARTS. 
 
 Parts of an 
 
 cwt. 
 
 Parts of 
 
 cwt. 
 
 Parts of 
 
 2 qrs. is 
 
 i 
 
 281b. is 
 
 A 
 
 I41b. is 
 
 1 
 
 i 
 
 14 
 
 - i 
 
 7 
 
 161b. is - 
 
 - J- 
 
 8 - - 
 
 1 
 
 4 - 
 
 14 
 
 i 
 
 7 
 
 - i 
 
 3* - 
 
 8 - 
 
 TV 
 
 Of 501b. 25 
 
 i 
 
 Of 251b. 1 
 
 7 
 
 - TV 
 
 1 ^ 
 
 J i 
 
 
 Of 100. 12^ 
 
 
 10 
 
 i 
 
 
 10 
 
 TV 
 
 5 
 
 TV 
 
 
 5 
 
 ^V 
 
 
 
 
 cwt. 
 
 i 
 
 R 
 
HO TARE AND TRET. 
 
 8. In I29cwt. 3qrs. 161b. gross, tare 141b. per cwt., 
 what is the neat weight ? Ans. 113cwt. 2qrs. 17^1b. 
 
 4. In 25 barrels of figs, each 2cwt. Iqr. gross, tare 
 161b. per cwt., how much neat ? Ans. 48cwt. Oqrs. 24lb. 
 
 CASE III. 
 
 When tret is to be allowed with tare,. 
 
 RULE. Divide the suttle weight by 26, and the quo- 
 tient is the tret, which subtract from the suttle, and the 
 remainder is the neat weight. 
 
 EXAMPLES. 
 
 1. In 9cwt. 2qrs. 171b. gross, tare 371b. and tret as 
 usual, how much neat 1 
 
 Cwt. 
 
 qr*. 
 
 tb- 
 
 9 
 
 2 
 
 17 gross. 
 
 
 1 
 
 9 tare. 
 
 26)9 
 
 1 
 
 8 tare-suttle. 
 
 
 1 
 
 12^ tret. 
 
 8 
 
 3 
 
 234-4- Answer. 
 
 2. In 7 casks of prunes, each weighing 4cwt. gross, 
 tare 17^1b. per cwt. and tret as usual, how much neat, at 
 251b. to the qr. ? Ans. 22cwt. 21-^lb. 
 
 3. What is the neat weight of 3hhds. o sugar weigh- 
 ing as follows ; -the first 4cwt. 51b. gross, tare 731b. ; the 
 second 3cwt. 2qrs. gross, tare 561b. ; and the third 2cwt. 
 3qrs. 171b. gross, tare 471b. ; and allowing tret to each as 
 usual ; 251b. a qr. ? Ans. 8cwt. Iqr. 12lb. 
 
 4. What is the neat weight of 10 casks of raisins, each 
 weighing 3cwt. 2qrs., tare 141b. per cwt. tret as usual ; 
 and what will be the amount at $15 per cwt. ? 
 
 Ans. 29cwt. Iqr. 22 T yb. $441, 70cts. 6m.-f 
 
 CASE IV. 
 
 When tare % tret, and doff, are all allowed. 
 
 RULE. Deduct the tare and tret, as before, and divide 
 the suttle by 168, and the quotient is the cloff, which 
 subtract from the suttle, and the remainder is the neat. 
 
DOUBLE RULE OF THREE. Ill 
 
 EXAMPLES. 
 
 1. What is the neat weight of a hhd. of Tobacco, 
 weighing 15cwt. 3qrs. 201b. gross, tare 71b. per cwt. tret 
 and cloff as usual 1 Cwt. qrs. ft. oz. 
 
 71b. is T V)15 3 20 gross. 
 3 27 8 tare. 
 
 26)14 3 20 8 tare-suttle. 
 2 85+ tret. 
 
 168)14 1 12 3 tret-suttle. 
 9 9+ cloff. 
 
 14 1 2 10 Answer, nearly. 
 
 NOTE. Some say 21b. for every lOOlb. of tret-suttle 
 ought to be allowed, to make the weight hold good when 
 sold by retail ; instead of 21b. on every 3361b. 
 
 2. What is the value, at 5d. per lb., neat weight, of 
 26 chests of sugar, each, 9cwt. 2qrs. 15Jflb. gross, tare 
 131b. per cwt., tret as usual, and clofF 21b. on 3001b. ; 
 251b. a qr. ? Ans. <499 15s. 5d. 
 
 DOUBLE RULE OF THREE. 
 
 THE DOUBLE RULE OF THREE teaches to solve such 
 questions as require two or more statings in the Rule of 
 Three. In these questions there is always given an odd 
 number of terms as five, seven, or nine, &c. These are 
 distinguished into terms of supposition, and terms of de- 
 mand, the number of the former always exceeding that 
 of the latter by one, which is of the same kind with the 
 term or answer sought. 
 
 RULE. Write the term of supposition, which is of the 
 same kind with the answer, for the middle term. 
 
 Take one of the other terms of supposition, and one of 
 the demanding terms of the same kind with it ; then 
 place one of them for a first term, and the other for a 
 third, according to the directions given in the Rule of 
 Three. Do the same with another term of supposition, 
 and its corresponding demanding term ; and so on, if 
 
112 DOUBLE RULE OF THREE 
 
 there be more terms of each kind, writing the terms un- 
 der each other, which fall on the same side of the middle 
 term. 
 
 Multiply together all the terms in the first place, and 
 also all the terms in the third place. Then multiply the 
 latter product by the middle term, and divide the result 
 by the former product; and the quotient will be the an- 
 swer required. Or, take the two upper terms and the 
 middle term, in the same order as they stand, for the first 
 stating of a question in the Single Rule of Three ; then 
 take the fourth number resulting from the first stating, 
 for the middle term of a new stating in the above Rule, 
 and the two under terms of the Double Rule of Three 
 stating, in the same order as they stand, for the extreme 
 terms of the new stating ; and the fourth term resulting 
 from this new stating, will be the answer. 
 
 NOTE 1. The first and third terms of each line, if of 
 different denominations, must be reduced to the same de- 
 nomination. 
 
 2. After stating, and before commencing the opera- 
 tion, if one of the first terms, and either the middle term 
 or one of the last terms will exactly divide by one and 
 the same number, let them be divided, and the quotients 
 used instead of them ; which will much shorten the work, 
 Make trial with the first Example. 
 EXAMPLES. 
 
 1. How many men can complete a trench of 135 yards 
 long in eight days, provided 16 men can dig 54 yards in 
 6 days ? 
 
 54 yards) : ,6 men:: { ^ Jg* j : 
 8 days ) f. o days ) 
 
 432 810 
 
 16 
 
 Here 54-5-27=2. 135-4-27=5 
 8.5-2=4. And 6-2=3 4860 
 
 Then 2x4=8, and 16-=-S=2. 810 
 
 And 5x3=15, arid 15x2=30. 
 
 Therefore, the answer by 432)12960(30 men, Ans. 
 
 Note 2, is 30 men. 1296 
 
 
 
DOUBLE RULE OF THREE. 113 
 
 2. If a100 in one year gain ^6 interest, what will be 
 the interest of ,750 for 7 years ? 
 
 -*.'* 
 
 . 
 
 3. A farmer sells 204 dollars' worth of grain in 5 years, 
 when it sold at 60 cents per bushel ; what is it per bushel 
 when he sells 1000 dollars' worth in 18 years, if he sell 
 the same quantity yearly? Ans. 8lcts. 6 T 9 ^mills.+ 
 
 4. If 7 men can reap 84 acres of wheat in 24 days, how 
 many men can reap 1 00 acres in 1 days 1 Ans. 20 men. 
 
 5. If 6 men build a wall 20 feet long, 6 feet high and 
 4 feet thick, in 16 days ; in what time will 24 men build 
 one 200 feet long, 8 feet high and 6 feet thick T 
 
 Ans. 80 days. 
 
 24 men } f 6 men } 
 
 20 feet long f , > 200 feet long f . go , 
 
 6 feet high ( ' ' ) 8 feet high ( ' 
 
 4 feet thick ; (6 feet thick ) 
 
 6. An usurer put out 75 dollars at interest, and at the 
 end of 8 months he received for principal and interest* 
 79 dollars ; I demand at what rate per cent, he received 
 interest. Ans. 8 per cent. 
 
 7. If the carriage of 13cwt. Iqr. for 72 miles be Z 
 10s. 6d. what will be the carriage of 7cwt. 3qrs. for 112 
 miles ? Ans. ^2 5s. lid. 1 -jZ&qr. 
 
 S. If a family of 9 persons spend 450 dollars in 5 
 months, how much would be sufficient to maintain them 
 8 months, if 5 more were added to the family ? 
 
 Ans. 1120 dollars. 
 
 9.. What is the interest of 654 dollars for 164*days, at 
 6 per cent, per annum ? Ans. 17 dolls. 63ets. 1m.+ 
 
 10. If 248 men in 5 days of 1 1 hours each, dig a trench 
 230 yards long, 3 yards wide and 2 deep ; in how many 
 days of 9 hours long, will 24 men dig a trench 420 yards 
 long, 5 wide and 3 deep ? Ans. 28S^VV days. 
 
 11. If 30 men perform a piece of work in 20 days, how- 
 many men will effect another piece of work, 4 times as 
 large, in a fifth part of the time ? Ans, 600 men. 
 
 12. Wlmt principal will gain 315 in 7 years, at 6 per 
 cent, per annum ? Ans. $750. 
 
 13. If SOOOJfe. of beef will serve 340 seamen 15 days, 
 how many pounds will serve 120 seamen 25 days T 
 
 Ans. 1764ft. 
 
114 CONJOINED PROPORTION. 
 
 CONJOINED PROPORTION. 
 
 CONJOINED PROPORTION is when the coins, weights, or 
 measures of several countries, are compared in the same 
 question ; or it is the joining together of several ratios, 
 and inferring the ratio of the first antecedent and last 
 consequent, from the ratios of the several antecedents and 
 their respective consequents. 
 
 CASE I. 
 
 When it is required tojind liow many of the last kind of 
 
 coin, weight, or measure, mentioned in the question, are 
 
 equal to a given number of thejirst. 
 
 RULE 1. Multiply continually together the antecedents 
 for the first term, and the consequents for the secondi arid 
 make the given number the third. 
 
 2. Then find the fourth term or proportional, which 
 will be the answer required. 
 
 EXAMPLES. 
 
 1. If lOffe. at Boston make 9ffo. at Amsterdam ; 
 
 at Amsterdam 112j^. at Thoulouse, how many Jfo. at 
 Thoulouse are equal to 50}fc. at Boston 1 
 Ant. Con. 
 
 10 : 9 
 
 90 : 112 
 
 900 : 1008 : : 50 : 56ft. Ans. 
 
 2. If 20 brasses at Leghorn be equal to 10 varas at Lis- 
 bon ; 61 varas at Lisbon to 75 American yards; how 
 many American yards are equal to 100 brasses at Leg- 
 horn ? Ans. 61ff yards. 
 
 CASE II. 
 
 WJien it is required to jftnd how many of the Jirst kind of 
 coin, weight, or measure, mentioned in the question, are 
 equal to a given number of the last. 
 RULE. Proceed as in the first case, only make the pro- 
 
 duct of the consequents the first term, and that of the an- 
 
 tecedents the second. 
 
BARTER. 115 
 
 EXAMPLES. 
 
 1. If 20 ells English make 11 canes at Rome; 44 
 canes at Rome, 136 brasses at Venice ; how many ells 
 English make 85 brasses at Venice ? 
 
 Ant. Con. 
 20 : 11 
 44 : 136 
 
 880 1496 : 880 : : 85 : 50 ells, Ans. 
 
 2. If 41 U. S. bushels make 26 hanegas at Cadiz ; 39 
 hanegas at Cadiz, 162 alquiers at Lisbon ; 27 alquiers at 
 Lisbon, 5 sacks at Leghorn ; and 20 sacks at Leghorn, 
 21 tuns at Copenhagen ; how many U. S. bushels make 
 36 tuns at Copenhagen ? Ans. 70fbus. 
 
 3. If 300 U. S. miles make 77 miles in Germany ; 
 1771 miles in Germany, 1250 posts in France ; and 25 
 posts in France, 38 miles in Holland ; how many U. S. 
 miles make 80 in Holland ? Ans. 290| U. S. miles. 
 
 BARTER. 
 
 BARTER is the exchanging of one commodity for anoth 
 er, and directs traders so to proportion their goods, that 
 neither party may sustain loss. 
 
 RULE.* Find the value of that commodity the quan- 
 tity of which is given ; then find what quantity of the 
 other, at the rate proposed, you may have for the same 
 money, and it gives the answer required. 
 
 EXAMPLES. 
 
 1. How many dozen of candles, at 3s. 6d. per dozen, 
 must be given in barter for 4cwt. 2qrs. of tallow, at 46s. 
 per cwt. ? 
 
 * This rule is only an application of the Rule of Three. 
 
116 BARTER. 
 
 qrs. s. cwt. qrs. 
 
 4 : 46 : : 4 & 
 
 18 4 
 
 368 18 
 46 
 
 . : 4)828 
 2,0)20,7 
 
 s. d. doz. 
 
 36: 1 :: 
 
 2. A buys of B 4 hogsheads of wine containing 410 
 gallons, at 1 dollar 17 cents per gallon; and 253 {fo. of 
 coffee at 21 cents per ft- : In part of which he pays him 
 21 dollars in cash, and the balance in boards at 8 dollars 
 per thousand ; how many feet of boards does the balance 
 require ? Ans. 63978J feet. 
 
 3. Bought a sloop of 70 tons at 16 dollars per ton; 
 paid in cash 500 dollars, 350 gallons of molasses at 64cts. 
 per gallon, and the balance in oil at 74 cts. per gallon ; 
 how many gallons did it amount to? 
 
 Ans. 535-gygaI. 
 
 4. A barters with B 150 bushels of wheat at 5s. 9d. 
 per bushel, for 65 bushels of corn at 2s, lOd. per bushel, 
 and the balance in oats at 2s. Id. per bushel ; what quan- 
 tity of oats must A receive ? Ans. 325f bushels. 
 
 5. How much wine at 1 dollar 28 cents per gallon, 
 must I receive in barter for 26cwt. 2qrs. 14j^. of raisins, 
 at 9 dollars 44 cents 4 mills per cwt., 25}^. a qr. ? 
 
 Ans. 196gals. 2qts. l,704gills. 
 
 6. A delivers B 3 hogsheads of wine at 6s. Sd. per 
 gallon, for 126 yards of cloth ; what was the cloth per 
 yard, in Federal Money T Ans. 1 dollar 66f cts. 
 
 7. A has a quantity of pepper* weight neat 1600}^. at 
 Is. 5d. per Jfo. which he barters with B for two sorts of 
 goods, the one at 5d. the other at 8d. per pound, and to 
 have ^ in money, and of each sort of goods an equal 
 quantity ; how many ffe. of each must he receive, and 
 how much in money ? 
 
 Ans. I394fflb. of each,, and .37 15s. 6d, 
 
LOSS AND GAIN. 117 
 
 8. A and B barter; A has 145 gallons of oil, at 
 $l,20cts. per gallon ready money, but in barter he will 
 have $l,35cts. per gallon; B has linen at 58 cents per 
 yard ready money ; how must B sell his linen per yd., in 
 proportion to A's barter price, and how many yards are 
 equal to A's oil ? 
 
 . ( B's linen is 65cts. 2m. barter price, and 
 5 ' ( he must give A 300yds. for his oil. 
 
 9. K and L barter ; K has woollen cloth worth $1,33 
 cts. per yard, which he barters at $l,54cts. with L., for 
 linen cloth at 50cts. per yard, which is worth 43cts. per 
 yard ; who has the advantage in barter, and how much 
 linen does L give K for 70 yards of woollen 1 
 
 . ( 215fyds. of linen, and L has the advantage, his 
 5 * ( proportional barter price being only 49^J cts. 
 
 10. J. Tucker and Jonathan Olmstead barter ; the for- 
 mer gives the latter 90 gallons of wine at $1,28 per 
 gallon ; for which the latter gives the former 10 guineas 
 at 28s. each, in money, and 500fJ). of cotton ; what is it 
 valued at per pound ? Ans. 13f| cts. 
 
 11. Giles Jackson has 100 reams of paper, at $1,33 
 cts. ready money, which in barter he sets down at $l,66f 
 cts. Robert Howard, sensible of this, has pamphlets at 8 
 cts. apiece ready money, which he adequately charges, and 
 insists, besides, on O f the price of those he parts with, 
 in money ; what number of the books is he to deliver in 
 lieu of Jackson's paper ? what cash will make good the 
 difference 1 and how much is Howard the gainer by this 
 affair? Ans. 1600 books to be delivered; $4l,6(3fcts. 
 Howard is to have in cash ; and the gain to Howard is 
 $41,<56fcts. 
 
 LOSS AND GAIN. 
 
 Loss AND GAIN is a rule that discovers what is gained 
 or lost in buying or selling goods; and instructs mer- 
 chants and traders to raise or lower the price of their 
 goods, so as to gain or lose a certain sum per cent. 
 
 Questions in this rule are performed by the Rule of 
 Three. 
 
118 . LOSS AND GAIN. 
 
 There is, indeed, great variety in questions in this 
 rule ; but they may be all easily solved by a little consid- 
 eration, and the following proportion, viz. That the 
 gains or losses, are in proportion as the quantities of goods. 
 
 EXAMPLES. 
 
 1. Bought 30 hogsheads of molasses for 600 dollars ; 
 paid in duties 20 dollars 66 cents, freight 40 dollars 
 78 cents, for storage 6 dollars 5 cents, and for insurance 
 30 dollars 84 cents : If I sell it at 26 dollars per hogs- 
 head, how much shall I gain per cent. ? 
 
 $. cts. 
 
 600 $26 
 
 20 66 30 hhds. 
 
 40 78 
 
 6 05 $780 00 Sold for. 
 
 30 84 698 33 Cost. 
 
 698 33 81 67 Gain. 
 
 $. cts. $. cts. $. $. cts. 
 
 698 33 : 81 67 :: 100 : 11 69+Ans. 
 
 2. At 3s. 6d. profit on the pound, how much per cent. ? 
 
 Ans. .17 10s. 
 
 3. If !{. of coffee cost 12cts. and it sold for 15 cents, 
 what is the profit on 2931b. neat ? Ans. Sdolls. 79cts. 
 
 4. If a gallon of wine cost 6s. 8d. and is sold for 7s. 
 2d. what is the gain per cent. ? Ans. 7 per cent. 
 
 5. Sold a repeating watch for 175 dollars, upon which 
 I lost 17 per cent, whereas I ought to have gained 20 per 
 cent. ; how much was it sold for under its just value ? 
 
 Ans. 78dolls. lct.+ 
 
 6. If I buy broadcloth for 13s. 5d. per yard, how must 
 I sell it to gain at the rate of 25 per cent. ? 
 
 . . s. d. s. d. 
 
 100 : 125 : : 13 5 : 16 9f Ans. 
 Or thus, 4)13 5 
 
 16s. 9|d. 
 
 7. Bought oil for 90 cents per gallon ; at what rate 
 must it be sold to gain 20 per cent. 1 Ans. 108cts. 
 
FELLOWSHIP. 119 
 
 8. Bought 115 gallons of wine at 1 dollar 10 cents per 
 gallon ; how many gallons of water must be put in, so as 
 to gain 5 dollars by selling it at 1 dollar per gallon 1 
 
 Ana. 16 gallons. 
 
 9. Bought a hhd. of molasses containing 119 gallons* 
 at 52cts. ger gallon ; paid for carting the same $l,25cts. ; 
 and by accident 9 gallons leaked out ; at what rate per 
 gallon must I sell the remainder, so as to gain $13 in the 
 whole ? Ans. 69cts. 2m. + 
 
 10. Bought llcwt. of sugar, at 6d. per ft- but could 
 not sell it again for any more than b 16s. per cwt. ; did 
 I gain or lose by, my bargain 1 Ans. Lost <2 11s. 4d. 
 
 11. Bought cloth at 17s. 6d. per yard, which on exam- 
 ination, I find to be much damaged ; and am, therefore, 
 content to lose 15 per cent, by it ; how must I sell it 
 per yard ? Ans. 14s. 10|d. 
 
 12. By selling broadcloth at $3,25cts. per yard, I lose 
 at the rate of 20 per cent. ; what is the prime cost of said 
 cloth per yard ? Ans. $4,6 cents, 2m. 
 
 13. If, when I sell cloth at $7 per yard, I gain $10 per 
 cent. ; what will be the gain per cent, when it is sold for 
 $8 per yard ? Ans. 825,71 cents, 4m.-f 
 
 14. If I sell a cwt. of sugar for $8, and thereby lose 
 12 per cent. ; what shall I gain or lose per cent, if I sell 
 4cwt. of the same sugar for $36 1 
 
 Ans. I lose only 1 per cent. 
 
 FELLOWSHIP. 
 
 FELLOWSHIP is a rule by which merchants, &c. trad- 
 ing in company with a joint stock, determine each per- 
 son's particular share of the gain or loss in proportion 
 to his share in the joint stock. 
 
 By this rule a bankrupt's estate may be divided among 
 his creditors ; as also legacies adjusted when there is a 
 deficiency of assets or effects. 
 
 SINGLE FELLOWSHIP. 
 
 Single Fellowship is when different stocks are employ- 
 ed for any certain equal time. 
 
120 SINGLE FELLOWSHIP. 
 
 RULE.* As the whole stock is to the whole gain or 
 loss, so is each man's particular stock to his particular 
 share of the gain or loss.f 
 
 PROOF. Add all the shares together, and the sum will 
 be equal to the gain or loss, when the work is right. 
 
 EXAMPLES. 
 
 1. A and B gained by trade ,182. A put into stock 
 .300 and B 400 ; what is each person's share of the 
 profit ? 
 
 . 
 300+400=700 : 182 : : 300 : 78 A's share. 
 700 : 182 : : 400 : 104 B's share. 
 
 182 Proof. 
 
 2. A man dying, bequeathed his estate to his three sons 
 in the following manner, viz. to the eldest he gave 1840 
 dollars, to the second 1550 dollars ; and to the third 960 
 dollars ; but it was found his whole estate was no more 
 than 1840 dollars ; what is each one's proportion 1 
 
 ( $77S,29!f the first. 
 Ans. < 655,63^f the second. 
 ( 406,06jf the third. 
 
 3. A and B companied ; A put in 450 dollars, and re- 
 ceived f of the gain ; what did B put in 1 Ans. $300. 
 
 4. Three merchants freight a ship with wine ; A load- 
 ed 110 tuns, B 97 tuns, and C 133 tuns. In a storm the 
 seamen were obliged to throw 85 tuns overboard ; how 
 much must each sustain of the loss 1 
 
 Ans. A 27 , B 24, C 33 tuns. 
 
 5. Three men, A, B, and C, contract to build the hull 
 of a vessel for 625 dollars ; A works 100 days, and his 
 work is estimated at 1 dollar 80 cents per day ; B works 
 101 days, estimated at 1 dollar 60 cents per day ; and C 
 
 * That the gain or loss in this rule is evidently in proportion to 
 their stocks, may be shown from the nature of the Rule of Three. 
 
 f The first and third contractions of note 4 of the Rule of 
 Three, are often the best for working- questions in this ruley 
 especially in decimals. 
 
SINGLE FELLOWSHIP. 121 
 
 work 98 days, at 1 dollar 50 cents per day : how much 
 is each man's proportion according to his work? 
 
 day. $. cts. days. day. $. cts. days. day. $. cts. days. 
 1 : 1 80 :: 100 1 : I 60 : : 101^ * ^ 
 
 100 1 
 
 180,00 A's work. 
 162,00 B's do. 
 147,00 C's do. 
 
 147,00 
 489 162,00 
 
 $. $. $. $ cts. 
 
 489 : 625 : : 162 : 207,05^ 
 489 : 625 : : 147 : 187,88 T % 
 $. $. $. $ cts. 
 
 489 : 625 : : 180 : 230,06^ 
 
 ( $230,06-^ A's share. 
 AnsJ 207,05 T VvB's do. 
 ( 187,8S T 5 e% C's do. 
 
 $625,00 Proof. 
 
 6. A ship worth 3600 dollars being entirely lost, of 
 which ^ belonged to A, to B, and the rest to C ; what loss 
 will each sustain ? Ans. A $450. B $900. C $2250. 
 
 7. A and B gained 1260 dollars of which A is to have 
 ten per cent, more than B : what is the share of each ? 
 
 Ans. A 660dolls. B 600dolls. 
 
 8. Three merchants made a joint stock A put in ^565 
 6s. 8d. B ,478 5s. 4d. and C a certain sum ; they gained 
 <3739s. lid. of which C took ^112 Us. lid. for his 
 part ; what is A and B's part of the gain, and how much 
 did C put in ? ( A's gain 141 6s. 8d. 
 
 AnsJ B's do. 119 Us. 4d. 
 ( C put in 150 7s. 8d. 
 
 9. A, B, and C, traded in company ; A put in $140 ; 
 B, $250 ; and C put in 120yds. of cloth, at cash price ; 
 they gained $230, of which C took $100 for his share of 
 the gain ; how did C value his cloth per yard in com- 
 mon stock, and what was A's and B's part of the gain ? 
 
 A ( C put in his cloth at $2 per yard ; B's part of 
 5t \ the gain was $83,33^cts. ; and A's 5546,66f cts, 
 mt 
 
 , 
 
122 DOUBLE FELLOWSHIP. 
 
 DOUBLE FELLOWSHIP. 
 
 DOUBLE FELLOWSHIP is when the stocks are employed 
 for different times. 
 
 RULE.* Multiply each man's stock by the time of its 
 continuance ; then say, as the sum of all the products is 
 to the whole gain or loss, so is each man's particular pro- 
 duct to his particular share of the gain or loss. 
 EXAMPLES. 
 
 1. A and B hold a piece of ground in common, for 
 which they pay 36 A put in 23 oxen for 54 days, B 21 
 oxen for 70 days ; what part of the rent must each man pay 1 
 -23x54 = 1242 
 21x70=1470 
 
 1470 : 19 10 3 B's. 
 
 .36 Proof. 
 
 2. Two merchants enter into partnership for 16 months 
 A put in at first 1200 dollars, and at the end of 9 months 
 200 dollars more ; B put in at first 1500 dollars, and at 
 the expiration of 6 months took out 500 dollars with 
 this stock they gained 772 dollars 20 cents ; what is each 
 man's part of it 1 
 
 Ans. A's $401 70cents B's $370 50 cents. 
 
 3. A, B and C, enter into partnership ; A put in 85 
 dollars for 8 months, B put in 60 dollars for 10 months, 
 and C 120 dollars for 3 months; by misfortune they lost 
 41 dollars ; what part of the loss must each man sustain ? 
 
 Ans. A's part $17. B's $15. C's 9. 
 
 4. W. Thomas and N. White were joint tenants of a 
 mill, in the building of which Thomas laid out g!50, and 
 White S270. At the end of 7 months, Thomas sold his 
 share to White ; and at the end of the first year White 
 sold the mill. They then made a settlement; and, the 
 year's profit of the mill being ascertained at $260, what 
 was each man's share ? 
 
 Ans. Thomas's $54,16|cts. White's- $205,83 Jets. 
 
 * When the times are equal, the shares of the gain or loss are 
 evidently as the stocks, as in Single Fellowship ; and when the 
 stocks are equal the shares are as the times ;, but when neither 
 are equal, the shares must be as their products. 
 
SIMPLE INTEREST. 123 
 
 5. Jacob M'Ewen, Giles Jackson, John Hastings, and 
 Anthony Minot, were joint tenants of a certain toll bridge, 
 which they held for the terra of 14 years, by charter. 
 Their whole expense in building the bridge, was $'25745, 
 50cts., of which M'Ewen paid $489(>,67cts., Jackson 
 $1675, Hastings $12392,87cts., and Minot $6?8Q,96cts. 
 At the end of 2 years, M'Ewen sold out to Peter Thom- 
 son ; at the end^of 5 years, Jackson sold out to Jeremiah 
 Apthorp ; and at the end of 10 years, Hastings sold his 
 share to James Hawkins ; at the expiration of the 14 
 years, the whole tollage amounted to $30,000. What 
 was each man's share 1 
 
 'M'Ewen's, $10l8,90c. 2m.+ 
 Jackson's, G97,07c.+ 
 Hastings', I03l4,87c.+ 
 
 Answer. 
 
 Minot's, 7901,52c. 8m.+ 
 Thomson's, 4686,95c. 2m.+ 
 Apthorp's, 1254,72c. 6m.+ 
 Hawkins', 4125,94c. 8m. + 
 
 SIMPLE INTEREST. 
 
 INTEREST is the sum paid by the borrower to the lend- 
 er for the use of money lent. 
 
 The legal interest in most of the United States is 6 
 per cent, per annum ; that is, 6 for the use of <lOO, or 
 $6 for the use of $100 for one year, &c. 
 
 Principal is the money for which the compensation is 
 made. 
 
 Rate is the sum per cent, agreed on. 
 
 Amount is the principal and interest added together. 
 
 Interest is of two kinds ; simple and compound. Simple 
 Interest is that which is allowed for the sum lent only. 
 
 RULE.* Multiply the principal by the rate, and divide 
 the product by 100; and the quotient is the interest for 
 one year. Multiply the interest for one year by the given 
 number of years, and the product is the interest for that 
 time. For any parts of a year, as months, days, &c. di- 
 
 * Simple Interest is only an r application of the Rule of Three. 
 
124 
 
 SIMPLE INTEREST. 
 
 vide the interest for one year by the aliquot parts of a 
 year or month ; or the interest may be found by a state- 
 ment in the Rule of Three.* 
 
 NOTE. In Federal Money the quotient after the divis- 
 ion by 100 gives the answer in the same name with the 
 lowest denomination in the principal. 
 
 TABLE OF ALIQUOT PARTS. 
 
 Parts of a Year. 
 6 Months is - - 
 4 - - - 
 
 3 - - 
 
 2 - - 
 
 Parts of a Month. 
 
 3 Days is 
 
 o - 
 
 6 - - 
 10 - - 
 15 * - 
 
 TV 
 * 
 * 
 * 
 
 EXAMPLES. 
 
 1. What is the interest of J6639 for one year, at 6 per 
 cent.? Ans. ^38 6s. 9d.+ 
 
 639 
 6 
 
 qr. 2,40 
 
 * Calling 30 days a month gives the interest too much. If the 
 principal be small, the errour is trifling-. But, if the sum be 
 very large, say as, 365 days : is to the interest for 1 year : so 
 is the given number of days ; to me interest required. 
 
SIMPLE INTEREST. 125 
 
 2. What is the interest of 372 dollars for one year, 5 
 months, and 5 days, at 6^ per cent. ? 
 372 
 
 2232 
 186 
 
 4 months, $24,18 Interest for one year. 
 
 1 month, T L 8,06 Interest for four months. 
 
 5 days, | of 1 mo. 2,01^ Interest for one month. 
 
 do. five days. 
 
 834,59+ Answer. 
 
 3. What is the amount of <49 6s. 4^d. for one year, 
 at 6 per cent. ? 
 
 . s. d. 
 49 6 4 
 
 2,95 18 3 
 
 20 
 
 s. d. 
 
 19,18 49 6 4 Principal. 
 
 12 2 19 2 Interest. 
 
 2,19 Ans. =52 5 6 Amount. 
 
 4. What is the interest of 1600 dollars for one year 
 and three months, at 6 per cent. ? Ans. $120. 
 
 5. What is the interest of <71 7s. 6d. for 2 years, at 
 6 per cent. ? Ans. 8 11s. 3^d.+ 
 
 6. What is the interest of 67 dollars, 62 cents for 3 
 years and 2 months, at 6 per cent. ? 
 
 Ans. $!2,84cts. 7miHs.+ 
 
 7. How much is the interest of 325 dollars for 3 years, 
 at 6 per cent. ? Ans. $58, 50cts. 
 
 8. How much is the Jnterest of 66 cents, 4 mills for 
 one year and 7 months, at 6 per cent. ? 
 
 Ans. 6cts. 3mills.+ 
 
 9. What is the interest of 48 dollars, 25 cents, 5 mills 
 for 5 years, at 5 per cent. ? Aris* $12,6cts. 3mills.+ 
 
 L 2 
 
SIMPLE INTEREST. 
 
 10. What is the interest of 48 dollars for 3 years, at 9 
 percent.! Ans. $12,96cts. 
 
 11. What is the interest of 5 16s. 3d. for one year 
 and 6 months, at 6 per cent. ? Ans. 10s. 5jd.+ 
 
 12. What is the interest of 9672 dollars, from Nov. 
 30, 1823, to July 4, 1826, at 8 per cent. ? 
 
 Ans. $2007,42cts. 7ns. + 
 
 To find the interest of any sum at 6 per cent, per annum, 
 for any number of months. 
 
 RULE.* Multiply the principal by half the number of 
 months, and that product divided by 100 will be the inter- 
 est for the given time. 
 
 EXAMPLES. 
 
 1. What is the interest of 64 dollars, 50 cents for 8 
 months^ at 6 per cent. ? 
 
 64,50 
 
 8-^-2=4 4 
 
 1,00)2,58,00 Ans. 2 dolls. 58 cents, 
 
 2. How much is the interest of 36 dollars, 84 cents for 
 5 months, at 6 per cent. ? Ans. 92cts. 1 mill. 
 
 3. How much is the interest of 750 dollars for 15 
 months, at 6 per cent. 1 Ans. $56,25cts. 
 
 4. What is the interest of 24 15s. 4d. for 10 months, 
 at 6 per cent. I Ans. ~l 4s, 9d. y^qr. 
 
 5. What is the interest of 468 dollars for one month, 
 at 6 per cent, 1 Ans. $2,34cts. 
 
 To find the intertst of any sum for any number of days 
 
 when the rate is 6 per cent. 
 
 RULE. Multiply the principal by the number of days r 
 and divide the product by 6083 ; the quotient will be the 
 interest required. 
 
 * THE REASON OF THE RULE. When the time is months, mul- 
 tiplying- by the rate for the time gives the answer. 
 
 This rate is found by multiplying- the time by the given rate 
 per cent, for a year, and dividing- the product by 12 ; the quotien-t 
 is the rate required, and is always equal to half the months, when 
 the yearly rate is 6 per cent 
 
SIMPLE INTEREST. 127 
 
 EXAMPLES. 
 
 1. What is the interest of 376 dollars, 20 cents for 80 
 days, at 6 per cent. 1 376,20 
 
 80 
 
 $ cts. m. 
 
 6083)30096,00(4,94 7+ Answer. 
 
 2. What is the interest of 749 10s. 6d. for 12 days, at 
 6 per cent. 1 . s. d. 
 
 749 10 6 
 12 
 
 6083)8994 6 ()(<! 9s. 6fd.+ Answer. 
 
 SHORT PRACTICAL RULES 
 
 For calculating Interest at 6 per cent, either for months, 
 or months and days. 
 
 I. FOR POUNDS, SHILLINGS, &C. 
 
 RULE. If the principal consists of pounds only, cut off 
 the unit figure, and as it then stands, it will be the inter- 
 est, for one month, in shillings and decimal parts. ^Mul- 
 tiply this sum by the given number of months, and take 
 parts for the days, and you will have the answer in shil- 
 lings and decimal parts, which you may reduce. But if 
 the principal consist of pounds, shillings, &c. reduce it 
 to its decimal value ; then remove the decimal point one 
 place or figure, further towards the left hand, and as the 
 number then stands, it will show the interest for one 
 month, in shillings and decimals of a shilling. Multiply 
 by the given time, &c. and you will have the answer. 
 
 EXAMPLES. 
 
 1. Required the interest of ,48 for 5 months and 10 
 days, at 6 per cent. 
 Days. s. 
 
 !0||4, 8 interest for 1 month* 
 o 
 
 24, do. for 5 nionths. 
 1,6 do. for 10 days, 
 
 Ans. 25, 6 shillings =1 5s. 
 12 
 
 7,2 
 
128 SIMPLE INTEREST. 
 
 2. Required the interest of 56 10s. for 11 months, at 
 6 per cent. 
 
 . s. . 
 
 56 JO 56,5 decimal value. 
 Therefore 5,65 shillings interest for one month, 
 11 
 
 Ans. 62,15 interest for 11 months==3 2s. l,8d. 
 
 3. What is the interest of .37 8s. 6d. for one year, 7 
 months, and a half, at 6 per cent, per annum ? 
 
 Ans. 8 10s. 4d. 2,3qrs. 
 
 4. Required the interest of 28 16s. for 20 days, or 
 two-thirds of a month, at 6 per cent. Ans. Is. ll,04d. 
 
 NOTE. The foregoing rule will serve also, when the 
 rate is 5 per cent, or 7 per cent. For, when, by the Rule, 
 you have found the interest of any sum, at 6 per cent, if 
 you decrease the said interest by one-sixth of itself, it 
 then becomes the interest of the same sum, at 5 per cent. ; 
 or if, after finding the interest, by the Rule, at 6 per 
 cent., you increase the said interest by one-sixth of itself, 
 it then becomes the interest of the same sum, at 7 per cent. 
 
 Take for examples the first two of the preceding ones. 
 
 Required the interest of ^48 for 5 months and 10 days, 
 at 5 per cent. s. 
 
 J)25,6 interest at 6 per cent. 
 4,26 
 
 Ans. 21,33s. int. at 5 per cent=^l Is. 4d. 
 What is the interest of <56 10s. for 11 months, at 7 
 per cent. ? 
 s. 
 
 )62,15 int. at 6 per cent. 
 10,3583 
 
 Ans. 72,5083 sbill. int. at 7 per cent.=3 12s. 6, Id. 
 
 II. For Federal Money. 
 
 RULE. Divide the Principal by 2, placing the separa- 
 trix as usual, and the quotient will be the interest for one 
 month, in cents, and decimals of a cent ; that is the 
 
SIMPLE INTEREST. 129 
 
 figures at the left of the separatrix will be cents and 
 those on the right, decimals of a cent. Then multiply 
 this sum by the given number of months, or months and 
 decimal parts thereof, or for the days take even parts of 
 30, &c. 
 
 EXAMPLES. 
 
 1. Required the interest of $582,64cts. for 5^ months, 
 at 6 per cent. 
 
 2)562,64 
 
 281, 32 int. for 1 month. 
 
 140,66 int. for month. 
 1406,60 int. for 5 months. 
 
 Ans. 1547,26= l547 T 2^cts.$l5 3 47cts. 2,6m. 
 Or thus : 281,32 int. for 1 month* 
 X5,5 months. 
 
 1406 60 
 14066 
 
 1547,260cts.=$l5,47cts. 2,6m. 
 
 2. What is the interest of $l8,48cts.' for 3 years, 7 
 months, and 10 days, at 6 per cent. ? 
 
 2)18,48 
 
 10 days=i 9,24 int. for 1 month. 
 43 months. 
 
 2772 
 3696 
 
 897,32 int. for 43 months. 
 3,08 do. for 10 days. 
 
 Ans. 400,4,0cts.=$4 and 4 mills. 
 
 3. What is the interest of $468 for 7 months, at 6 per 
 cent.? Ans. $l6,38cts. 
 
 For 5 or 7 per cent, proceed according to the direc- 
 tions in the preceding Note. 
 
130 INTEREST ON NOTES, &,C. 
 
 To compute Interest on Notes, Bonds, fyc., having par- 
 tial Payments or Endorsements thereon. 
 
 RULE 1. Cast the Interest upon the whole Principal, 
 for the whole Time ; then separately upon each Endorse- 
 ment, for its respective Time ; and subtract the whole 
 Amount of the one from that of the other. 
 
 NOTE. I call 30 days a month, and 12 months a year. 
 
 EXAMPLES. 
 
 1. A Note dated January 1st, 1821, was given for 
 $120, interest at 6 per cent, and there were the follow- 
 ing payments endorsed upon it : 
 
 June 1, 1821, rec'd on the within note $50. 
 Oct. 1, 1821, rec'd $40. 
 
 I demand how much was due on said note, Jan'y 1, 1822, 
 when it was taken up ? 
 
 $ 
 
 120 Principal, or sum of the note. 
 7,20cts. Interest for the whole time. 
 
 $127,20cts. Amount. 
 
 9. 8. 
 
 50 First payment. 40 Second payment. 
 
 l,75cts. Interest. 0,60cts. Interest. 
 
 51,75 Its Amount. 40,60 Its Amount. 
 
 9 . c. 9. c. 
 
 51,75 ) Several Am'ts 127,20 Amount of Note. 
 
 40,60 } of Payments. 92,35 Amount of Pay'ts. 
 
 92,35 Total Amount. $34,85cts. Due on the Note, 
 
 Jan. 1, 1822, Ans. 
 
 2. A Bond dated April 17th, 1817, was given for $675, 
 interest at 6 per cent., and it was endorsed as follows : 
 
 May 7, 18:8, rec'd on the within $148. 
 
 August 17, 1820, rec'd $341. 
 
 Jan. 2, 1822, rec'd $99. 
 
 I demand what -was due on said bond, June 17, 1822, 
 when it was paid up ? Ans. $2l9,52cts. 
 
INTEREST ON NOTES, &C. 131 
 
 3. A bond or note was given, June 4th, 1819, for g699, 
 interest at 6 per cent. ; it was endorsed $78, July 9, 1820: 
 
 $147, Nov, 27, 1821; $6S, Jan. 17, 1822; and $400, 
 Jan, 30, 18*23 ; I demand how much will be due on said 
 note May 30, 1823? Ans. $l32,87cts. 1m. 
 
 62 10s. 
 
 4. Value received, I promise to pay George Appleton, 
 the sum of Sixty-two Pounds, Ten Shillings, and interest 
 at 6 per cent, per annum, till paicl. 
 
 PETER FRISBIE. 
 
 Halifax, April 4, 1820. 
 
 Endorsed, 50, Sept. 4, 1822. 
 
 If he endorse j(2l2 10s. June 1, 1823, how much will 
 be due on said note, December 4, 1823? 
 
 Ans. 9 12s. 
 
 Contraction of the Rule of 6 per cent. 
 
 RULE. Point off the right hand figure of each princi- 
 pal for a decimal ; multiply each by its particular time, 
 and add the products. If the principals be pounds, &c. 
 the sum total will be shillings and decimals ; but if Fed- 
 eral Money, halve it for dimes, interest: which added to 
 the last principal, gives the sum due on the note. 
 
 Take tliejirst Example in the preceding Rule. 
 
 1st Principal, $12,&x5 months. =60,0 
 2d do. 7,0x4 =28,0 
 
 3d do. 8,0x3 =09,0 
 
 As it is Federal Money, divide by 2)97,0 
 
 * 48,5dim.=48& 
 
 dimes=$4,85cts. which added to the last principal, 
 30+4,85 ~34,S5cts. amount due. Aris. 
 
 This is merely the same Rule as that at pages 128 129. 
 
132 INTEREST ON NOTES, &C. 
 
 For an Example in pounds, fye. take the last question in 
 the preceding Rule. 
 
 1st Principal 6% 10s.=6,25s. by Rule, page 127. 
 Yrs. m. d. 
 1822 9 4 
 1820 4 4 
 
 250 12 10 =2d Principals 1,25s. 
 
 12 
 
 Yrs. m. d. 
 
 29 months. 1823 6 
 s. m. s. 1822 9 4 
 
 Int. for Imo.=6,25x29 = 181,25 m. 
 
 Int. for Imo.=l,25x8,9= 11,125 8 27=8,9 
 
 192,375 shillings, 
 which =9 12s. 4d. Ans. 
 
 There being no principal left to which to add the above 
 sum, that sum is itself the exact answer. 
 
 This is the same Rule as that at page 127. 
 
 Note. If the last principal be overpaid by the last 
 endorsement, take the sum thus overpaid, multiply it, in 
 the same way, by the time, from that endorsement, up to 
 the settlement, then add its dimes, interest, to itself, and 
 subtract the sum from the interest of the Note's princi- 
 pals, as above found. 
 
 3. Gave a note for $500, Jan. 16, 1822, payable Nov. 
 16, 1826 ; $50 were endorsed thereon April 1, 1823; 
 $400, July 16, 1824; $(iO, Sept. 1, 1825; how will the 
 balance be at the date when payable 1 
 
 $500 = 1 st principal. 500 50=450=2d principal. 
 450 400=50=3d principal. Then the next endorse- 
 ment overpays the sum borrowed. 60=3d endorsement, 
 and 60 50, the 3d principal =$10 overpaid on the bum 
 borrowed. 
 
INTEREST ON NOTES, &C. 133 
 
 mo. mo. 
 
 1st prin. $50,0x14,5=725,0 Overp'd $1,0x26,5=26,5 
 2d do. 45,0x15,5=697,5 And 2)26,5 
 3d do. 5,0x13,5= 67,5 
 
 - Dimes 
 2)1490,0 Add sum 10,00 
 Then, $74,50 
 
 1 J /J&j- subtr. 745 dimes, $ 1 1 ,32 
 
 or, 74,50ct.s. -- 
 Aus. 863, 17 due on Note. 
 
 4. A note was given, May 10, 1821, for $685, interest 
 at 6 percent. Nov. 15, 1821, it was endorsed $150 ; April 
 30, 1822, g275 ; Dec. 1, 1822, $90 ; July 30, 1823, S45; 
 March 10, 1824, gf50, and Aug. 20, 1825, it was paid up ; 
 what was the last payment] Aris. $l6,96^cts. 
 
 RULE 2. The Supreme Court of Massachusetts, estab- 
 lished the following more equitable Rule. When there 
 are endorsements, find the interest on the principal to 
 the time of the first payment, add it to the principal, 
 and from the sum subtract the payment then made ; if 
 the endorsement be not equal to the interest then due, 
 cast the interest to the next endorsement, add the two 
 endorsements together, and from the amount of the prin- 
 cipal to that time, subtract their sum ; on the remainder 
 cast the interest to the time of the next endorsement, 
 subtracting each payment as you proceed, if it be not 
 less than the interest then arisen on the principal sum ; 
 and so on to the date of giving up the obligation. 
 
 EXAMPLES. 
 
 I. A note was given, January 20th, 1821, for $360, 
 SOcts; September 10th, following, 200 dollars were en- 
 dorsed, and December 20th, 1822, it was taken up ; what 
 was the last payment, interest at 6 per cent. ; computing 
 it by both rules, and showing their difference ? 
 Yr. m. d. Yr. m. d. 
 
 1821 9 10=Sep. 10. 1822 12 20=Dec 20. 
 
 1821 1 20=Jan. 20 1821 9 10=Sept 10. 
 
 7 20 1 3 10 
 
 M 
 
134 INTEREST ON NOTES, &C. 
 
 $. Cts. 
 
 360,50 Note dated Jan. 20, 1821. mo. d. 
 13,81 9 Interest up to Sept. 10, 1821=7 20 
 
 374, 31 9 Amount. 
 
 200, 00 First payment deducted. 
 
 174, 3i 9 Due Sept. 10, 1821. Yr. m. 
 
 13, 36 4 Interest to Dec. 20, 1822=1 3 
 
 $187, 68 3 Last payment by Rule 2. 
 
 Yr. m. d. 
 
 1822 12 20 
 
 1821 1 20 
 
 1 11 
 
 $. cts. 
 
 360, 50 Note. Yr. m. 
 
 41, 45 7 Interest up to Dec. 20, 1822=1 11 
 
 401, 95 7 Amount for the whole time. 
 
 200,00 First payment, Sept. 10, 1821. Y. m. d. 
 15, 33 3 Interest up to Dec. 20, 1822=1 3 10 
 
 215,33 3 Amount. 
 
 $. cts. m. 
 
 401, 95 7 Amount of note. 
 215, 33 3 Amount of payment. 
 
 186, 62 4 Last payment by Rule 1. 
 
 $. cts. m. 
 
 187, 68 3 Due by Massachusetts' Rule. 
 186, 62 4 Due by common Rule. 
 
 1, 05 9 Difference. 
 
 2. A note was given, Nov. 15, 1820, for g282, 56cts. ; 
 May 10, 1821, g96,34cts. were endorsed ; December 20, 
 following, 174,28ets. ; May 10, 1822, $lO,50cts.; and 
 November 15, following, $ 5,25cts. ; what will be due on 
 
COMPOUND INTEREST. 135 
 
 said note June 10, 1823, reckoning interest at 6 per cent., 
 computing by both rules, and showing their difference 1 
 
 f By Massachusetts' rule $13, 164 
 Answer. J By ru , e firgt 
 
 Difference $1, 709 
 
 3. An obligation was given, July 15, 1817, for $340, in- 
 terest at 6 per cent., on which are the following payments ; 
 Dec. 25, following, $7,50cts. ; June 10, 1818, $iO,25cts.; 
 January 1, 1819, $15; August 16, 1820, 825,75cts. ; 
 November 1, 1821, $30 ; and December 25, 1822, 250 ; 
 if it be taken up March 25, 1823, what sum will cancel 
 it, by the Massachusetts' rule ? Ans. $113, 53cts. 
 
 N. B. The figures beyond mills, in the two preceding 
 examples, are all omitted in casting. 
 
 4. If a note, given July 5, 1 819, for $101, at 6 per cent., 
 were endorsed July 5, 1820, 86 ; July 5, 1821, 6; July 
 5, 1822, 6; July 5, 1823, $6; July 5, 1824, $6; July 
 5, 1825, 86; July 5, 1826, $6; July 5, 1827, $6; July 
 5, 1828, $6; what would be due, July 5, 1829, by both 
 rules and how much more by the second rule than by the 
 first ; 
 
 r$ 107,60 by Mass, rule; $91,40 by common 
 Ans. < rule ; and 816,20 more by the former than the 
 ( latter. 
 
 COMPOUND INTEREST. 
 
 COMPOUND INTEREST is what arises from the interest 
 being added to the principal, and becoming a part of the 
 principal, at the end of each stated time of payment. 
 
 RULE. Find the Simple Interest of the given sum for 
 one year, or the time of the first payment ; add it to the 
 principal, and find the interest of the amount for the next 
 year or payment, and so on for the number of payments 
 required. Subtract the principal from the last amount, 
 and the remainder will be the compound interest. 
 
136 COMPOUND INTEREST. 
 
 EXAMPLES. 
 
 1. What is the compound interest of 406 dollars, for 3 
 years, at 6 per cent, per annum ? 
 
 406 principal for the 1st year. 
 6 
 
 24,36 interpst of do. > 
 
 40(5, principal for the 1st year. } 
 
 430,36 principal for the 2d year. 
 6 
 
 25.82,1,6 interest of do. ) , 
 
 430,36 principal for the 2d year. / * 
 
 456,18,1 principal for the 3d year. 
 6 
 
 27,37,0,86 interest for do. > A 
 
 456,18,1 principal for the 3d year. J A 
 
 483,55,1 amount for three years. 
 
 406, principal for the 1st year, subtracted. 
 
 Ans. $77,55,1 compound interest. 
 
 2. How much is the compound interest of 2535 dollars 
 for four years, at 6 per cent, per annum ? 
 
 Ans. 665 dolls. 36cts.+ 
 
 N. B. The mills are here all rejected in casting. 
 
 3. What is the compound interest of 1000 dollars for 5 
 years, at 6 per cent. ? Ans. 338dolls. 22cts. 4mills.+ 
 
 N. B. The figures beyond mills are here omitted in 
 casting. The next is done in whole numbers. 
 
 4. What is the compound interest of ^128 17s. 6d. for 
 6 years, at 6 per cent. 1 Ans. .53 18s. 8d.+ 
 
 5. How much will 680 dollars amount to in 4 years, at 
 6 per cent, compound interest ? 
 
 Ans. S858,48cts. 3m. + 
 
 N. B. Figures beyond mills omitted. 
 
COMMISSION. INSURANCE. 137 
 
 COMMISSION.* 
 
 COMMISSION AND BROKERAGE are compensations to 
 factors and brokers for their respective services. 
 
 EXAMPLES. 
 
 1. What is the commission on 4760dolls. at 2J- pe-r 
 cent. ? 
 
 2)4760 
 
 9520 
 2380 
 
 119,00 Ans. 119 dollars. 
 
 2. What is the commission on ,526 11s. 5d. at 3J- per 
 cent.? Ans. IS 8s. 7d.+ 
 
 3. What is the brokerage on 926 dollars, 59 cents, at 
 1 per cent. 1 Ans. 13dolls. 89cts. 7m.-J- 
 
 4. What is the commission on 1293 dollars, 53 cents, 
 at I per cent. 1 Ans. 9dolls. 73cts. 8mills,+ 
 
 5. Required the neat proceeds of certain goods 
 amounting to 2176 dollars, deducting a commission of - 
 per cent. ? Ans. 2i5(idolls. 96cts. 
 
 6. A factor receives 3690 dollars to lay out in potash, 
 reserving from it his commission of 2^ per cent, on the 
 purchase; the potash being 190 dollars per ton, how 
 much did he purchase ? Ans. 18tons, IScwt. 3qrs. 22 T 2 F ft. 
 
 INSURANCE. 
 
 INSURANCE is an exemption from hazard, by paying a 
 certain sum on condition of being indemnified for loss or 
 damage of ships, houses, merchandise, &c. which may 
 happen from storms, fires, &c. 
 
 * The method of working 1 questions in this and the following 
 rules of Insurance, &c. is the same as in Simple Interest. 
 M2 
 
138 DISCOUNT. 
 
 EXAMPLES. 
 
 1. What is the premium of insuring 8250 dollars, at 6 
 per cent. 1 
 
 8250 
 6 
 
 495,00 Ans. 495 dollars. 
 
 What is the premium of insuring 1650 dollars, at 
 per cent. ? Ans. 255dolls. 75cts. 
 
 3. What sum must be received for a policy of 1658 
 dollars, deducting a premium of 23 per cent, for insur- 
 ance ? Ans. 1276dolls. 66cts. 
 
 4. What is the premium for the insurance of 4000 dol- 
 lars, at 7f per cent. ? Ans. 305 dollars. 
 
 5. What sum must be insured upon to cover 1800 
 dollars, when the premium is 10 per cent. ? 
 
 100 Policy. 
 Deduct 10 Premium. 
 
 90 Sum covered. 
 If $90 : $100 : : $1800 : $2000 Ans. 
 
 DISCOUNT. 
 
 DISCOUNT is an allowance made for the payment of any 
 sum of money before it becomes due ; arid is the differ- 
 ence between that sum due some time hence, and its 
 present worth. The present worth of any sum, due some 
 time hence, is such a sum, as, if put to interest, would in 
 that time, and at the rate per cent, for which the discount 
 is to be made, amount to the sum or debt then due. 
 
 RULE. As the amount of 100 dollars for the given rate 
 and time, is to the interest of 100 dollars for that time, so 
 is the given sum or debt, to the discount required. Or, 
 as the amount of 100 dollars or pounds, is to 100, so is 
 the given sum or debt, to the present worth required. 
 
 NOTE. When goods are bought or sold, money ad- 
 vanced, bank-bills exchanged, &c. and discount is to be 
 
DISCOUNT. 139 
 
 made at any rate per cent, without time, the interest of 
 the sums as found for a year, is the discount. 
 
 EXAMPLES. 
 
 1. What is the discount of 1912 dollars, 50 cents due 
 3 years hence, at 4 per cent. 1 
 
 4,50 
 3 
 
 13,50 
 100, 
 
 g. cts. $. cts. 8. cts. 
 
 $113,50 : 13,50 : : 1912,50 : 227,47+Ans. 
 
 2. What is the present worth of 760 dollars due in 8 
 months, discount at 6 per cent, per annum ? 
 
 8. 
 
 8 mo. 46=4 
 100 
 
 8. 8. 
 
 104 : 100 :: 760 : 
 
 Ans. 730dolls. 76cts. 9mills.+ 
 
 3. what is the present worth of 500 dollars payable in 
 ^ of a year, discount being at 5 per cent. 1 
 
 Ans. $493,S2fcts.+ 
 
 NOTE. When several sums are to be paid at various 
 times, find the discount or present worth of each sum sep- 
 arately, and then add those discounts of present worths 
 into one sum, in order to obtain the required answer. 
 
 4. A is to pay 592 dollars, 70 cents on the first day of 
 April, 1833, and 598 dollars, 90 cents the first of July 
 following. It is required to know how much money will 
 discharge both sums on the first of January, 1833, dis- 
 counting at 8 per cent, per annum. 
 
 Ans. H56dolls. 94cts. 3mills.+ 
 
 5. Bought a quantity of goods for 500 dollars ready 
 money, and sold them again for 666 dollars, 67 cents, pay- 
 able at f of a year ; what was the gain in ready money, 
 supposing discount to be made at 5 per cent. ? 
 
 Ans. 142dolls. 57cts.+ 
 
 6. How much ready money will discharge a note for 
 150 dollars due in 60 days, allowing 6 per cent, per an- 
 num discount! Ans. 148dolls. 51cts. 4mills.+ 
 
140 ANNUITIES. 
 
 7. If a legacy of 2000 dollars be left to me ; 500 dol- 
 lars payable in 6 months ; 800 in one year ; and the rest 
 at the end of 3 years ; and the executor be willing to 
 make me present payment, discounting at 6 per cent. ; 
 what ought I to receive 1 Ans. 183;3dolls. 37cts. 4m. + 
 
 8. What is the present worth of 60, payable at 3 
 and 6 months, at 5 per cent, per annum discount ? 
 
 Ans. c58 17s. lid. 
 
 ANNUITIES. 
 
 AN ANNUITY is a yearly income arising from money, 
 &c. and is either paid for a term of years, or upon a 
 life. Annuities or pensions are said to be in arrears, 
 when they are payable or due either yearly, half-yearly, 
 or quarterly, and yet remain unpaid for any number of 
 payments. 
 
 The sum of all the annuities, for the time they have 
 been forborne, together with the interest due upon each, 
 is called the amount. 
 
 If an annuity be bought off, or paid all at once at the 
 beginning of the first year, {he price which is paid for it, 
 is called the present worth. 
 
 CASE I. 
 
 To find the amount of an annuity at Simple Interest. 
 
 RULE. 1. Find the interest of the given annuity for 
 1 year; and then for 2, 3, <fcc. years, up to the given 
 time less 1. 2. Multiply the annuity by the number of 
 years given, and add the product to the whole interest, 
 and the sum will be the amount required. 
 
 EXAMPLES. 
 
 1. If 250 dollars, yearly annuity, be forborne 7 years, 
 what will it amount to in that time, allowing simple in- 
 terest at 6 per cent, per annum ? 
 
 1st. Interest of $250, at 6 per cent, for 1 year=$lo. 
 2yrs. = f30. 3yrs.=^$45. 4yrs.=:$60. 5yrs.=f 75, and 
 6yrs.=$90. And 2d. $250x7 = $1750. 
 
 Then, 15+30+45+60+75+90+1750=$2065 Ans, 
 
EQUATION OF PAYMENTS. 141 
 
 2. A house is leased for 7 years, at 400 dollars per an- 
 num ; and the rent is unpaid during the whole term; 
 what sum is due at the end of the lease, simple interest 
 being allowed, at 6 per cent, per annum 1 
 
 Ans. 0304 dollars. 
 
 CASE II. 
 
 To find the present worth of an annuity at Simple Interest. 
 
 RULE. Find the present worth of each year by itself, 
 
 discounting from the time it falls due ; the sum pf all 
 
 these present worths, will be the present worth required. 
 
 NOTE. The divisions are continued decimally. But 
 it is deemed more equitable to allow compound interest, 
 in purchasing annuities. 
 
 1. What is the present worth of 400 dollars per annum, 
 to continue 5 years, at 6 per cent, per annum ? 
 
 106 ") f 377,35849 ^present worth of 1 yr. 
 
 112 I |357,14285= 3d yr. 
 
 118 V: 100 : : 400 :{ 338,98305= 3d yr. 
 
 124 j I 322,580ti4~ 4th yr. 
 
 130 j [307,6923 5th yr. 
 
 Ans. 1703,75733 = $1703,75cts. 7m. 
 
 2. What is a salary of $300 per annum, to continue 
 5 years, worth, in ready money at 5 per cent, per ann. ? 
 
 Ans. $1309,31cts.+ 
 
 3. What is ,250 yearly rent, to continue 6 years, 
 worth in ready money, at 3 per cent. ? 
 
 Ans. 1380 7s. 10f AD- 
 
 EQUATION OF PAYMENTS. 
 
 EQUATION OF PAYMENTS is finding a time to pay at 
 once several debts due at different times, so that no loss 
 shall be sustained by either party. 
 
142 EXCHANGE. 
 
 RULE. Multiply each payment by the time at which 
 it is due ; then divide the sum of the product by the sum 
 of the payments, and the quotient will be the time re- 
 quired. 
 
 EXAMPLES. 
 
 1. A owes B 1900 dollars, to be paid as follows, viz. 
 500 dollars in 6 months, 600 dollars in 7 months, and 800 
 dollars in 10 months; what is the equated time to pay 
 the \vhole debt 1 
 
 500 x 6=3000 
 
 600 x 7=4200 
 
 800x10=8000 
 
 1900 ) 15200(8 months, Ans. 
 15200 
 
 2. A owes B 240 dollars to be paid in six months ; but 
 in 1^ months pays him 60 dollars, and in 4 months after 
 that 80 dollars more ; how much longer than six months 
 should A in equity defer the rest 1 Ans. 2 T 7 y months. 
 
 3. I owe 6512 dollars, to be paid ^ in 3 months, in 5 
 months, in ten months, and the remainder in 14 months ; 
 at what time ought the whole to be paid ? 
 
 Ans. 6 months. 
 
 4. A owes 60 dollars to be paid in 90 days, 75 dollars 
 in 60 days, and 50 dollars in 30 days ; what is the equat- 
 ed time for the whole to be paid 1 Ans. 61 T 6 TF days.+ 
 
 EXCHANGE. 
 
 BY this rule merchants know what sum of money ought 
 to be received in one country, for a given sum of differ- 
 ent specie paid in another, according to a given course of 
 exchange. 
 
 By the following table the moneys of foreign nations 
 may be reduced to that of the United States. 
 
EXCHANGE. 143 
 
 TABLE 
 
 Showing the value of the coins or moneys of account, 
 of foreign nations, estimated in Federal Money, agreea- 
 bly to a law of the United States in most instances. 
 
 8 cts. 
 
 Pound Sterling of Great Britain, 4 44 
 
 Pound Sterling of Ireland, - - 4 10 
 
 Livre of France, - 18 
 
 Guilder, or Florin, of Holland, - 39 
 
 Marc Banco of Hamburgh, - - 33^ 
 
 Rix Dollar of Denmark, and Germany, 1 00 
 Rial Plate of Spain, 10 
 
 Milre of Portugal, - - - 1 24 
 
 Tale of China and Japan, 1 48 
 
 Pagoda of India, - - 1 94 
 
 Rupee of Bengal, - 55 
 
 Ruble of Russia, - 1 00 
 
 Testoon of Italy, - 23f 
 
 Pound English W. I. Islands, - 3 16f 
 
 Livre French W. I. Islands, - 
 
 Franc of France, - 
 
 Rix Dollar of Sweden, . - 1 
 
 Do. do. of Austria, - 
 
 Do. do. of Poland and Prussia, - 
 
 Medio of Morocco, - - J 
 
 Piaster of Arabia, - 1 00 
 
 Or of Persia, - - - 1 48 
 
 I. Of Great Britain. 
 EXAMPLES. 
 
 1. in ^45 10s. sterling, how many dollars and cents ? 
 A pound sterling being =444 cents. Therefore, 
 
 as 1 : 444cts. : : 45,5 : 20202cts.=$202,02cts. Ans. 
 
 2. In $500 how many pounds sterling? 
 
 As 444cts. : 1 : : 50000cts. : ,112 12s. 3d.+ Ans. 
 
 NOTE. This method is not so exact as that given in 
 Reduction of Currencies. 
 
144 EXCHANGE. 
 
 2. Of Ireland. 
 
 1. In ,85 7s. 6d. Irish money, how many cents ? 
 
 1 Irish=410cts. 
 
 . cts. . cts. $ cts. 
 
 Therefore, as 1 : 410 : : 85,375 : 35,003 J=350,03 
 
 2. In $176,30cts., how many pounds Irish ? 
 
 As 410cts. : 1 : : 17630cts. : <43 Irish. Ans. 
 
 3. Of France. 
 
 Accounts are there kept in Hvres, sols, and deniers. 
 f 12 deniers, or pence, make 1 sol, or shilling. 
 ( 20 sols, or shillings 1 livre or pound. 
 
 EXAMPLES. 
 
 1. In 360 livres, 12 sols, how many dollars and cents ? 
 
 1 livre, of France = 18cts. or 185mills. 
 I. m. L m. 
 
 As 1 : 185 :: 360,6 : 667 11 =$66,7 lets. 1m. Ans. 
 
 2. What sum of French money is equal to $94,52cts. 
 8m. t m. liv. m. liv. so. den. 
 
 As 185 : 1 : : 94528 : 510 19' 2+ Ans. 
 
 4. Of Holland, 8fc. 
 
 Accounts are kept there in guilders, stivers, grotes, 
 and peningens. 
 
 C 8 peningens make 1 grote'or penny. 
 < 2 grotes 1 stiver. 
 
 ( 20 stivers 1 guilder or florin. 
 
 1 guilder =39ets. =390 mills. 
 
 EXAMPLE. 
 
 In 250 guilders, 16 stivers, how much Federal Money ? 
 
 guil. cts. guiL $. d. c. m. 
 
 As 1 : 39 : : 250,8 : 97 8 1 2 Ans. 
 
 m. g. m. g. 
 
 As 390 : 1 : : 97812 : 250,8 Proof. 
 
 5. Of Hamburgh, in Germany. 
 
 Accounts are kept in Hamburgh, in marcs, shillings 
 and fennings, and by some in rix dollars. 
 
EXCHANGE. 145 
 
 / 12 fennings make 1 shilling lub, 
 J 16 shillings - 1 marc. 
 ( 3 marcs - 1 rix dollar. 
 
 1 marc=33cts. or one third of a dollar. 
 
 RULE 1. Divide the marcs by 3, and the quotient will 
 be dollars. 
 
 EXAMPLE. 
 
 Bring 584 marcs, 12 shillings to Federal Money. 
 
 3)584,75 
 
 $194,9lc. 6m. Ans. 
 
 RULE 2. When the given sum is Federal Money, mul- 
 tiply it by 3, and the product will be marcs and decimal 
 parts. 
 
 EXAMPLE. 
 
 In $3l5,70cts. how much Hamburgh, &c. money ? 
 315,70 
 3 
 
 947,10=947 marcs, 1 shil, 7, 2 fennings. Ans. 
 
 6. Of Spain. 
 
 In Spain, they keep their accounts in piastres, rials, 
 and maravedies. 
 
 {34 naaravedies of plate make 1 rial of plate. 
 8 risfts of plate make 1 piastre, or piece of 8. 
 1 rial =10 cents, or 1 dime. 
 
 RULE 1. To reduce rials to Federal Money, point off 
 the rigfit hand figure of the given sum, and as it stands, 
 it is the answer in dollars and dimes. 
 
 EXAMPLE. 
 
 In 968 rials, how many dollars, &c. ? 
 
 Ans. 96,8 ^=$96,8 dimes, or $96, SOcts- 
 
 RULE 2. To reduce cents into rials, divide the cents 
 by 10. 
 
 N 
 
146 EXCHANGE. 
 
 EXAMPLE. 
 In $94,65cts. bow many rials, &c. ? 
 
 9465 10 =946,5 =946 rials, 17 maravedies. Ans. 
 
 7. Of Portugal 
 
 In Portugal, accounts are kept in niilrez and rez, reck- 
 oning 1000 rez to a milre, as its name imports. 
 1 milre=z:l24 cents. 
 
 RULE 1. To reduce milrez into Federal Money, mul- 
 tiply the given sum by 124, and the product will be cents ; 
 if the given sum contain rez, multiply as before, and cut 
 off 3 figures from the right of the product ; the answer 
 will be cents, and decimals of a cent. If the rez are less 
 than 100, prefix a cipher to them, before you annex them 
 to the milrez. A comma, or decimal point, is used to 
 separate rez and milrez. 
 
 EXAMPLES. 
 
 1. In 580 milrez how many cents ? 
 
 580x124=71920 cents = $719,20cts. Ans. 
 
 2. In 125 milrez, 96 rez, how many cents ? 
 125,096 x 124 = 1551 l,904cts. or $l55,llcts. 9m.+Ans. 
 
 RULE 2. To reduce cents into milrez, divide the sum 
 by 124, and if decimals arise, continue the division to 
 three places of decimals ; the whole numbers in the quo- 
 tient will be milrez, and the decimals will be rez. 
 
 EXAMPLES. 
 
 1. How many milrez in 6878 cents ? 
 
 6878124=55,467+ or 55milr. 467 rez. Ans. 
 
 2. In $31,09cts3 mills, how many milrez of Portugal ? 
 
 c. m. 
 3109,3-;-124=25,075=25milr. 75rez. Ans. 
 
 8. Of East-India Money. 
 
 RULE. To reduce India Money to Federal. Multiply 
 tales of China by 148; pagodas of India by 194; and 
 rupees of Bengal, by 55J : and the product, in every case, 
 will be cents. To change Federal Money into the before- 
 
VULGAR FRACTIONS. 147 
 
 named moneys, divide the given sum by one of the above 
 numbers, as the case may be. 
 
 EXAMPLES. 
 
 1. In 1000 tales of China, how many dollars ? 
 
 Ans. $1480. 
 
 2. In 1420 tales of China, how many dollars, &,c. ? 
 
 Ans. 2101, 6Gcts. 
 
 3. In $948,68cts. how many tales of China ? 
 
 Ans. 641. 
 
 4. In 482 pagodas of India, how many dollars and 
 cents 1 Ans. 8935,8cts. 
 
 5. In S135,80cts. how many pagodas of India? 
 
 Ans. 70. 
 
 6. In 920 rupees of Bengal, how much Federal Money ? 
 
 Ans. $510,60cts. 
 
 7. In $54,39cts. how many rupees of Bengal ? 
 
 Ans. 98. 
 
 A compound Example in European Money. 
 
 Suppose a merchant of Hallowell has effects at London, 
 to the amount of $5460, which he can remit by way of 
 Lisbon, at 1 milre per dollar, and thence to Boston, at 
 $l,50cts. per milre ; or by way of Paris, at 5 livres per 
 dollar ; thence to Brest, at 6 livres per crown, and thence 
 to Washington, at $I,25cts. per crown; which will be 
 the most advantageous way of remitting ? 
 
 A f Remitting by way of Lisbon is most 
 3 * \ advantageous by $2502,50cts. 
 
 VULGAR FRACTIONS. 
 
 VULGAR FRACTIONS were briefly introduced immediate- 
 ly after Duodecimals ; and some general definitions, and 
 a few such problems as were necessary to prepare the 
 scholar for, and to lead him into decimals, were there given. 
 Those general definitions he is requested to read anew. 
 
148 VULGAR FRACTIONS. 
 
 Vulgar Fractions are either proper, improper, single, 
 compound or mixed. 
 
 1. A proper fraction is when the numerator is less 
 than the denominator : as, f , f, f, &,c. which mean of 
 1, f of l,f of l,&c. 
 
 2. An improper fraction is when the numerator ex- 
 ceeds the denominator : as, f , -y T -, &c. 
 
 3. A single fraction is a simple expression for any 
 number of parts of the integer. 
 
 4. A compound fraction is the fraction of a fraction : 
 as, of f, | of , &c., which mean of f of l,f of of 
 1, &c. 
 
 5. A mixed number is composed of a whole number 
 and a fraction ; as, 8-J, 12 T 9 7j, &,c. 
 
 NOTE. Any number may be expressed like a fraction 
 by writing 1 under it : Thus -f- means 6 ones or 6. 
 
 A fraction having a fraction or mixed number for its 
 numerator or denominator, or both, is called a complex 
 fraction. A fraction denotes division, and its value is 
 equal to the quotient obtained by dividing the numerator 
 by the denominator : thus -^ is equal to 3, and ^- equal 
 to 4. Therefore, if the numerator be less than the de- 
 nominator, the value of the fraction is less than 1. If 
 the numerator be the same as the denominator, the frac- 
 tion is just equal to 1. And if the numerator be greater 
 than the denominator, the fraction is greater than 1. 
 
 6. The common measure of two or more numbers is 
 that number which will divide each of them without a 
 remainder ; and the greatest number that will do this, is 
 called the greatest common measure. 
 
 7. A number which can be measured by two or more 
 numbers, is called their common multiple ; and if it be the 
 least number, which can be so measured, it is called their 
 least common multiple. 
 
 PROBLEM 1. 
 
 To find the greatest common measure of two or more 
 numbers. 
 
 RULE. 1. If there be two numbers only, divide the 
 greater by the less, and this divisor by the remainder, ancl 
 
VULGAR FRACTIONS. 149 
 
 so on ; always dividing the last divisor by the last remain- 
 der, until nothing remains ; then will the last divisor be 
 the greatest common measure required. 
 
 2. When there are more than two numbers, find the 
 greatest common measure of two of them, as before ; and 
 next find the greatest common measure of that common 
 measure and one of the other numbers ; and so on, 
 through all the numbers to the last ; then will the greatest 
 common measure last found be the answer. 
 
 3. If one happen to be the common measure, the given 
 numbers are prime to each other, and found to be incom- 
 mensurable. 
 
 EXAMPLES. 
 
 1. Required the greatest common measure of 918,1998 
 and 522. 
 
 918)1998(2 
 
 1836 So 54 is the greatest common 
 
 measure of' 1998 and 918 
 162)918(5 Hence 54)522(9 
 
 810 486 
 
 108)162(1 36)54(1 
 
 108 36 
 
 54)108(2 18)36(2 
 
 108 36 
 
 Therefore 18 is the answer required. 
 
 2. What is the greatest common measure of 612 and 
 540? Ans. 36. 
 
 3. What is the greatest common measure of 117 and 
 91 1 Ans. 13. 
 
 PROBLEM 2. 
 
 To find the least common multiple of two or more numbers. 
 
 RULE. 1. Divide by any number, that will divide two 
 
 or more of the given numbers without a remainder, and 
 
 set the quotient, together with the undivided numbers, 
 
 in a line beneath. 
 
 2. Divide the second line as before, and so on, until 
 there are no two numbers that can be divided ; then the 
 N 2 
 
150 VULGAR FRACTIONS. 
 
 continued product of the divisors and quotient will give 
 the multiple required. 
 
 EXAMPLES. 
 
 1. What is the least common multiple of 3, 5, 8, and 
 101 
 
 2)3 5 8 10 
 
 5)3 545 
 
 3141 Then 2x5x3x4=120 Ans. 
 
 2. What is the least common multiple of 9, 8, 15, 16 ? 
 
 Ans. 720. 
 
 3. What is the least number that 3, 4, 8, and 12 will 
 measure ? Ans. 24. 
 
 4. What is the least number that can be divided by the 
 9 digits without a remainder Ans. 2520. 
 
 REDUCTION OF VULGAR FRACTIONS. 
 
 REDUCTION OF VULGAR FRACTIONS is the bringing 
 them out of one form into another, in order to prepare 
 them for the operations of addition, subtraction, &c. 
 
 CASE I. 
 
 To abbreviate or reduce fractions to their lowest terms. 
 
 RULE. Divide the terms of the given fraction by any 
 number that will divide them without a remainder, &c. 
 as in Rule of Problem 1, page 76. Or divide both the 
 terms of the fraction by their greatest common measure, 
 and the quotients, will be the terms of the fraction re- 
 quired. If a fraction have ciphers on the right hand of 
 both its terms, it may be reduced by cutting off an equal 
 number from both. 
 
VULGAR FRACTIONS. 151 
 
 EXAMPLES. 
 
 1. Reduce JM to its lowest terms. 
 
 (4) (3) (4) 
 
 it* = lf=iS=f the answer. 
 Or thus, 144)240(1 
 144 
 
 96)144(1 
 96 
 
 Therefore 48 is the greatest 
 48)96(2 common measure, and 
 96 48)^=f Ans. 
 
 NOTE. 1. Any number ending with an even number, 
 or a cipher, is divisible by 2. 
 
 2. Any number ending with 5 or 0, is divisible by 5. 
 
 3. If the right hand place of any number be 0, the 
 whole is divisible by 10. 
 
 4. If the two right hand figures, of any number are 
 divisible by 4, the whole is divisible by 4. 
 
 5. If the sum of the digits, constituting any number, 
 be divisible by 3 or 9, the whole is divisible by 3 or 9. 
 
 6. All prime numbers, except 2 and 5, have 1, 3, 7, or 
 9, in the place of units; and consequently all other num- 
 bers are composite, and capable of being divided. 
 
 7. When numbers with the sign of addition or subtrac- 
 tion between them, are to be divided by any number, 
 each of the numbers must be divided. Thus, 
 
 4+8+10 
 
 =2+4+5=11 
 
 2 
 
 8. But if the numbers have the sign of multiplication 
 between them, only one of them must be divided. Thus, 
 
 3x8x10 x3x4xlO 1x4x10 1x2x10 
 
 , m 2_0 fffl 
 
 2x6 1x6 1x2 1x1 
 
 9. If both the numerator and denominator of a frac- 
 tion be multiplied or divided by the same number, the 
 fraction will still retain its original value. 
 
 Let and T 9 ^ be two fractions proposed ; then 1x1=^; 
 and T 9 ^-rf=f' That is, if the numerator 4, and denom- 
 
152 VUGLAR FRACTIONS. 
 
 inator 5, of the first fraction, be each multiplied by the 
 same number 2, the produced fraction T 8 ^ is equal to the 
 proposed one f. For the numerator and denominator of 
 the produced fraction, are in the same proportion as the 
 numerator and denominator of the proposed one. Also, 
 if the numerator 9, and the denominator 12, of the second 
 fraction, be each divided by the same number 3, the frac- 
 tions J and T 9 2- are equal for the same reason. 
 
 2. Reduce -fe to its lowest terms. Ans, -j^-. 
 
 3. Reduce -f/-^ to its lowest terms. Ans. f. 
 
 4. Reduce T 4 T 9 g^ to its lowest terms. Ans. fff . 
 
 5. Reduce ff-Jf to its lowest terms. Ans. . 
 
 CASE II. 
 
 To reduce a mixed number to its equivalent improper 
 fraction. 
 
 RULE. Multiply the whole number by the denomina- 
 tor of the fraction, and add the numerator to the pro- 
 duct; then that sum written above the denominator, 
 will form the fraction required. 
 
 EXAMPLES. 
 
 1. Reduce 27|- to its equivalent improper fraction. 
 
 27 
 9 
 
 243 
 2 
 
 245 Or 27x9 +2 
 
 =-|^ the answer. 
 
 9 9 
 
 2 Reduce 514 T ^ to an improper fraction. Ans. ^ 
 
 3. Reduce I2|f to an improper fraction. Ans. 
 
 4. Reduce 79{-f to an improper fraction. Ans. 
 
 5. Reduce 100|| to an improper fraction. Ans. 
 
TULfSAR FRACTIONS. 153 
 
 CASE III. 
 
 To reduce an improper fraction to its equivalent whole or 
 
 mixed number. 
 
 RULE. Divide the numerator by the denominator, and 
 the quotient will be the whole or mixed number required. 
 
 EXAMPLES. 
 1. Reduce & to its equivalent whole or mixed Dumber. 
 
 19)081(61 A 
 
 90 
 
 21 
 16 
 
 5 or ^=98116=61^ the answer. 
 
 2. Reduce 5 r y 9 - to its equivalent whole or mixed num- 
 ber. Ans. 12ff . 
 
 3. Reduce *g- to its equivalent whole or mixed num- 
 ber. Ans. 2f. 
 
 4. Reduce -^ to its equivalent whole or mixed num- 
 ber. Ans. 7. 
 
 5. Reduce **$* to its proper terms. Ans. 1209|f J. 
 
 CASE IV. 
 
 To reduce a whole number to an equivalent fraction, hav- 
 
 ing a given denominator* 
 
 RULE. Multiply the whole number by the given de- 
 nominator, and place the product over the said denomi- 
 nator, and it will form the fraction required. 
 
 EXAMPLES. 
 
 1. Reduce 7 to a fraction, whose denominator shall be 
 9. 7x9=63, and & Answer. 
 
 And ^.=63-5-9=7 Proof. 
 
 2. Reduce 13 to a fraction, whose denominator shall 
 be 12. Ans. J^/. 
 
 3. Reduce 746 to a fraction, whose denominator shall 
 be 60, Ans. 
 
154 VULGAR FRACTIONS. 
 
 CASE V. 
 
 To reduce a compound fraction to an equivalent single one. 
 
 RULE. Multiply all the numerators together for the 
 numerator, and all the denominators together for the de- 
 nominator, and they will form the fraction required. 
 
 If part of the compound fraction be a whole or mixed 
 number, it must be reduced to an improper fraction by 
 one of the former cases. 
 
 When it can be done, divide any two terms of the 
 fraction by the same number, and use the quotients in- 
 stead thereof. 
 
 EXAMPLES. 
 
 1. Reduce of of f- of T 8 T to a single fraction. 
 2x3x4x8 
 
 =|-ff =|f the answer. Or, by expunging 
 
 3x4x5x11 
 
 equal numerators and equal denominators, the answer 
 will be as before, =^f. 
 
 2. Reduce f of f of f to a single fraction. Ans. %. 
 
 3. Reduce T ^ of T 7 7j of T 8 ^ of 10 to a single fraction. 
 
 Ans. 4/^-f- . 
 
 4. Reduce J- of f of f to a single fraction. Ans. 8^ 8 T . 
 
 CASE VI. 
 
 To reduce fractions of different denominators to equiva- 
 lent fractions, having a common denominator. 
 
 RULE. Multiply each numerator into all the denomi- 
 nators except its own, for a new numerator; and all the 
 denominators continually for the common denominator; 
 first reducing the fractions to their lowest terms, &c. 
 
 NOTE. By Note 9, in case 1, it will be seen, that sev- 
 eral fractions of different denominators may be readily 
 reduced to a common denominator. Thus -J- may be re- 
 duced to the same denominator as f by multiplying its 
 terms by 3, by which it becomes f-. Also ^-, f , and ^ 
 may be reduced to a common denominator, by multiply- 
 ing the terms of the first fraction by 6, of the second by 
 3, and dividing those of the last by 5. And so of others. 
 
VULGAR FRACTIONS. 155 
 
 EXAMPLES. 
 
 1. Reduce , f , and f- to equivalent fractions, having a 
 common denominator. 
 
 1 x5x7=35 the new numerator for i. 
 3x2x7=42 do. do. f. 
 
 4x2x5=40 do. do. f 
 
 2x5x7=70 the common denominator. 
 Therefore the equivalent fractions are f, tf an( * tS- 
 the answer. 
 
 2. Reduce T 6 <y, f, , and f to equivalent fractions, hav- 
 ing a common denominator. Ans. f |f , f~J$, ^$7, and -| f g. 
 
 3. Reduce , f of f , 5^ and T 2 ^ to a common denomi- 
 nator. Ans. #$, ff, Jy^, ^ 6 T V 
 
 4. Reduce |^, f of 1^, T 9 T and f to a common denom- 
 inator. Ans. iM 
 
 CASE VII. 
 
 To find the value of a fraction in any known parts of 
 
 the integer. 
 
 RULE. Multiply the numerator by the parts in the 
 next inferiour denomination, and divide the product by 
 the denominator ; and if any thing remain, multiply it 
 by the next inferiour denominator, and divide by the de- 
 nominator as before ; and so on, as far as necessary ; and 
 the quotients placed after one another, in their order, will 
 be the answer required. 
 
 EXAMPLES. 
 
 1. What is the value of f of a shilling ? Ans. 
 3 
 12 
 
 8)36(4d. 
 32 
 
 4 
 4 
 
 8)16(2qrs. 
 16 
 
156 VULGAR FRACTIONS. 
 
 2. What is the value of T 5 ^ of a dollar 1 
 
 Ans. 41cts. 6f mills. 
 
 3. What is the value of f of a mile ? 
 
 Ans. 4fur. 22pol. 4yds. 2|feet. 
 
 4. What is the value of f of a month ? 
 
 Ans. 3w. Id. 9h. 36m. 
 
 5. What is the value of T 7 F of an acre ? 
 
 Ans. Irood, SOpoles. 
 
 6. What is the value of f of f of of $49,95cts. 1 
 
 Ans. $5,55cts. 
 
 CASE VIII. 
 
 To reduce a fraction of one denomination to that of 
 
 another, retaining the same value. 
 
 RULE. Make a compound fraction of it, and reduce 
 it to a single one. 
 
 EXAMPLES. 
 
 1. Reduce f of a penny to the fraction of a pound. 
 
 of j of 2V=TA^=^F the answer. 
 And s-fg- of ^ of -y=f|f=|d. the proof. 
 
 2. Reduce ^ of a farthing to the fraction of a pound 
 
 Ans. Tsfoj. 
 
 3. Reduce - of anill to the fraction of a dollar. 
 
 4. Reduce -^ to the fraction of a penny. 
 
 Ans. -4_-= 
 
 5. Reduce f of a pound Avoirdupois to the fraction of 
 an cwt. Ans. T f =3f ? . 
 
 6. Reduce T % of a month to the fraction of a day. 
 
 Ans. ff 
 
 7. Reduce 7s. 3d. to the fraction of a pound. 
 
 5. d. s. 
 
 73 20 in a 
 
 12 12 
 
 87 numerator. 240 denominator. 
 
 8. Reduce 6 furlongs, 16 poles to the fraction of a mile. 
 
 Ans. f. 
 
VULGAR FRACTIONS. 157 
 
 ADDITION OF VULGAR FRACTIONS. 
 
 RULE. Reduce compound fractions to single ones ; 
 mixed numbers to improper fractions ; fractions of differ- 
 ent integers to those of the same ; and all of them to a 
 common denominator ; then the sum of the numerators, 
 written over the common denominator, will be the sum 
 of the fractions required. 
 
 NOTE 1. In adding mixed numbers that are not com- 
 pounded with other fractions, find first the sum of the 
 fractions, to which add the whole numbers of the given 
 mixed number. 
 
 NOTE 2. When adding fractions of money,weight,&c. 
 reduce fractions of different integers to those of the same 
 integer. Or, find the value of each fraction by Case 7, 
 in Reduction, and then add them in their proper terms. 
 
 EXAMPLES. 
 
 1. Add 3f , I, f of I and 7 together. 
 
 First, 3f=^-, f of =f = T ^, 7HF- 
 Then the fractions are 5 g 9 -, -|, T 7 <y and . 
 29x8 xiOx 1=2320 
 7x8xlOx 1= 560 
 7x8x 8x 1= 448 
 7x8x 8x10=4480 
 
 7808 
 
 =12|ff=12 
 
 8x8x10x1=640 
 
 2. Add f , 7, and -J of f together. Ans. S| . 
 
 3. What is the sum of of 95 and J of 14 ? 
 
 Ans. 43-|4. 
 
 4. Add 19, 7, and J- of f together. Ans. 26|. 
 
 5. Add f and 17^ together. Ans. 18 J. 
 
 6. What is the sum of } 9 f s. and T 5 ? of a penny ? 
 
 Ans. fi$fs. or~3s. Id. l^qrs. 
 
 7. Add $ of 15. 3f , of f of f of a pound and f 
 of f of a shilling together. Ans. 7 1 7s. 5fd. 
 
 8. Add f of a yard, of a foot and f of a mile to- 
 gether. Ans. 660yds. 2ft. 9in. 
 
 9. Add of a week, ^ of a day and of an hour to- 
 gether. Ans. 2days, 14hours. 
 
 O 
 
158 VULGAR FRACTIONS. 
 
 10. Add f- of a ton to $ of an hundred weight, 
 
 a qr. Ans. 12cwt. Iqr. 7ft. 13oz. llfdrs. 
 
 11. Suppose I have f of a ship worth $6000, and that 
 I buy another person's share of her, which is T \ ; what 
 part of her belongs to me then, and what is it worth 1 
 
 Ans. I have ft, and it is worth $4125. 
 
 SUBTRACTION OF VULGAR FRACTIONS. 
 
 RULE. Prepare the fractions as in addition, and the 
 difference of the numerators written above the common 
 denominator, will give the difference of the fractions re- 
 quired. 
 
 NOTE 1. In subtracting mixed numbers, when the 
 fraction in the subtrahend is greater than that in the min- 
 uend, subtract the numerator of the subtrahend from the 
 denominator, and to the difference add the numerator of 
 the minuend, and carry one to the integer in the subtra- 
 hend. If the minuend contain no fraction, proceed in 
 the same way, there being then nothing to add to the 
 difference. 
 
 NOTE 2. In fractions of money, weights, &c. you 
 may find the value of each of the given fractions, by Case 
 ?, in Reduction, and then subtract them in their proper 
 terms. 
 
 EXAMPLES. 
 
 1. From f take f of f 
 
 | of f=/ T 
 / T =f =4- the answer. 
 
 2. From ||J take . Ans. -f-f?. 
 
 3. From 7ll take ft. Ans. 70ff. 
 
 4. From < take f of a shilling. Ans. 16s. 9d. 
 
 5. From foz. take of a pennyweight. 
 
 Ans. lldwt. 3grs. 
 
 6. From cwt. take F of a pound. 
 
 Ans. Iqr. 27 tb- 6oz. lOfdrs. 
 
 7. From 7 weeks take 9 days T 7 ^. 
 
 Ans. 5w. 4d. 7h. 12m. 
 
 8. From 4 days 7 hours take 1 day 9 hours ^. 
 
 Ans. 2d. 22h. 20m. 
 
VULGAR FRACTIONS. 159 
 
 9, Suppose I own f of a farm, which is worth $3600, 
 and that I sell f of my share ; what part of it have I left 
 and what is it worth ] 
 
 Ans. fc ; and worth $750. 
 
 MULTIPLICATION OF VULGAR FRACTIONS. 
 
 RULE. Reduce compound fractions to single ones, 
 mixed numbers to improper fractions, and those of differ- 
 ent integers to the same ; then multiply all the numera- 
 tors together for the numerator, and multiply all the de- 
 nominators together for the denominator, of the product 
 required. But always, before multiplying, cancel equal 
 numerators and denominators, and divide those that are 
 divisible by the same numbers, both here and in division, 
 agreeably to what is seen in Note 8 of Case 1, in Reduc- 
 tion, page 150, and in Case 5, page 154. 
 
 NOTE. A fraction is best multiplied by an integer, by 
 dividing the denominator by it, if possible, but if that 
 cannot be done, multiply the numerator by it. 
 
 EXAMPLES. 
 
 1. Required the continued product of 2^, -J, , of f 
 and 2. 1x5 
 
 2JHb * of f = - =T\ and 2=f . 
 
 3x6 
 
 ^5x1x5x2 
 
 Then J XX T 5 F Xf = -- = T ^ Ans. 
 2x8x18x1 
 
 2. Multiply f by T 3 T . Ans. T V 
 
 3. Multiply 4 by \. Ans. T V 
 
 4. Multiply f of 8, by of 5. Ans. 21. 
 
 5. Multiply 7 by 9f Ans. 69f . 
 
 6. Multiply 4J-, |- of ^, and 184 continually together. 
 
 Ans. 9 T |^. 
 
 7. What is the product of 5, f , f , of f , and 4 ? 
 
 Ans. 
 
 DIVISION OF VULGAR FRACTIONS. 
 
 RULE. Prepare the fractions as in Multiplication ; 
 then invert the divisor, and proceed as in Multiplication. 
 The product will be the quotient required. 
 
160 RULE OF THREE IN VULGAR FRACTIONS. 
 
 NOTE. A fraction is divided by an integer, by dividing 
 the numerator by it, if possible, but if it will not exactly 
 divide, then multiply the denominator by it. 
 
 EXAMPLES. 
 
 1. Divide | of 19 by f of . 
 
 1x19 19 
 | of 19= -- = , and f of f = T 6 2-= ; 
 
 5x 1 5 
 19x2 
 
 =-^-=7f the quotient required. 
 
 5x1 
 
 2. Divide ^ by f. Ans. Iff. 
 
 3. Divide 99 by 108. Ans. A 9 ff =ii- 
 
 4. Divide f of f by O f f . Ans. l. 
 
 5. Divide 3 by 9. Ans. . 
 
 6. Divide by 4. Ans. ^ 
 
 RULE OF THREE IN VULGAR FRACTIONS. 
 
 RULE. Prepare the fractions as in Multiplication ; 
 and state the question as directed in the Rule of Three 
 in whole numbers ; then invert the first term in the stat- 
 ing, and multiply all the three terms continually together, 
 numerators by numerators and denominators by denomi- 
 nators, and the product will be the answer in the same 
 name as the second or middle term. 
 
 EXAMPLES. 
 
 1. If f of a yard cost T 7 ^ of a pound, what will T \ of 
 an ell English cost 1 
 
 3x4x1 
 First f of a yard=f of f of = --- =f of an el! - 
 
 5x1x5 
 
 Ell . Ell. 
 Then : & : : A - 
 And f|-X T Vx T \=f|-^=9s. 8fd. Answer. 
 
 NOTE. Here, 25 and 15 are divisible by 5, giving 5 
 and 3 for quotients ; and 6 and one of the 12s. are divisi- 
 ble by 6, giving 1 and 2 for quotients ; then the numerators, 
 are 5, 7, and 1, whose product is 05 ; and the denomina- 
 tors 12, 2, and 3, whose product is 72. And thus always 
 
RULE OF THREE IN DECIMALS. 161 
 
 proceed, in multiplying vulgar fractions, by cancelling 
 equal numerators and denominators, or dividing those 
 that are divisible by the same number. 
 
 2. If f of a yard cost of a dollar, what will 40f yds. 
 come to? Ans. $59,42cts. 7 T 1 2 -m. 
 
 3. If 50f bushels of wheat cost $59^, what is it per 
 bushel? Ans. $ 1,1 Gets. 6fm. 
 
 4. If I buy lOOffo. of butter for $10, how many hun- 
 dred can I have for $105fi? Ans. 10^ hundred. 
 
 5. If ^ of of a yard of lace cost 30cts., how much 
 is it per yard ? Ans. $2,40cts. 
 
 6. If f of a gallon of wine cost f of a dollar, what will 
 of a tun cost ? Ans. $140. 
 
 7. A younger brother received $2200, which was just 
 y 5 ^ of his elder brother's fortune ; and 3 times the el- 
 der's money was j- as much again as the father was 
 worth ; what was the father worth ? Ans. $11000. 
 
 8. When the days are 13f hours long, a traveller 
 performs his journey in 35 days ; in how many days 
 will he perform the same journey, when the days are 
 11 T 9 Q- hours long? Ans. 40f^f days. 
 
 9. A schoolmaster being asked how many scholars he 
 had, answered, If I had as many, and^- as many, and % 
 as many, I should have 99 ; how many had he ? Ans. 36. 
 
 RULE OF THREE IN DECIMALS. 
 
 RULE. Reduce vulgar fractions to decimals, and com- 
 pound numbers either to decimals of the higher names, 
 or to integers of the lower, as also the first and third terms 
 to the same name; then state the question and proceed 
 as in whole numbers ; the fourth term will be the answer. 
 
 EXAMPLES. 
 
 1. If f of a yard cost f of a pound, what will of an 
 ell English cost? 
 
 f=r,375yd. f =,4. ell= T Vyd.-,3125yd. 
 
 Yd. . Yd. . 
 
 As ,375 : ,4 : : ,3125 : ,333+ =6s. 8d. Ans. 
 02 
 
162 DOUBLE RULE OF THREE IN VULGAR TRACTIONS. 
 
 2. If Icwt. Iqr. 16f fc. of sugar cost $10,9cts., what 
 will 9cwt. 3qr. cost, at the same rate ? 
 
 As 1,4 : 10,09: : 9,75 : 70,269 + =$70,26cts.9m.+ Ans. 
 
 3. A man bought 5,8 tuns of oil, for $266, but by ac- 
 cident, lost 50,9 gallons ; how must he sell the residue 
 per gallon, so as not to lose in his bargain 1 
 
 Ans. I8cts. 8m. + 
 
 4. If 12 oxen graze down 16,25 acres of grass in 20 
 days, how much of the same pasture will suffice 24 oxen 
 100 days 1 Ans. 162,5 acres. 
 
 5. By how many men would 417,6 acres be mowed in 
 12 days, if 5 men mow -J- of that quantity in the time ? 
 
 Ans. 20 men. 
 
 6. Two men bartered ; A had 24,07 yards of linen, for 
 which B gave him 25,6 ells of Holland, at 45cts. per ell ; 
 what was the linen per yard ? Ans. 47cts. 8m. + 
 
 7. A man bought a piece of broadcloth for $S5,20cts., 
 at the rate of $4,26cts. per yard ; how many yards did 
 it contain 1 Ans. 20yds. 
 
 8. If, during tbe tide of ebb, a boat should set out 
 from Hallowell for Bath, and at the same instant another 
 should put off at Bath for Hallowell, taking the distance 
 by water at 25 miles ; the tide forwards one, and retards 
 the other, say 2J- miles an hour; the boats are equal in 
 size and form, and equally laden, the rowers equally good, 
 and in the common way of working, in still water, would 
 proceed at the rate of 5 miles an hour ; where in the riv- 
 er would the two boats meet ? 
 
 Ans. They will meet 18f miles from Hallowell. 
 
 9. Bought 12 pieces of Holland for 404$, at $1^ 
 per ell Flemish ; how many yards were in each piece ? 
 
 Ans. 22|yds. 
 
 DOUBLE RULE OF THREE IN VULGAR 
 FRACTIONS. 
 
 RULE. State the question as in the same rule in whole 
 numbers; invert the fractions which stand in and under 
 the first term ; then proceed as in the Rule of Three in 
 Vulgar Fractions* 
 
SIMPLE INTEREST BY DECIMALS. 
 
 163 
 
 EXAMPLES. 
 
 I. If 2 10s. be the wages of 15 men for 6 days, what 
 will be the wages of 12 men for 18 days 1 
 
 J^ men ) . 5 ^ , ( -^ men 
 f days J : ** I Af days ' 
 
 == 6 ^ 2s. 2d. 2fqrs. Ans. 
 
 NOTE. In this solution, having stated the question 
 properly, inverted the two fractions in the first term and 
 then arranged all the terms to be multiplied into each 
 other, I take the shorter method of cancelling equal nu- 
 merators and denominators, and abridging such as are di- 
 visible by the same number. The 12 cancels the 2 arid 
 the 6, and the 5 divides the 15, leaving a denominator 3 
 to be multiplied into the only remaining denominator, 
 which is 3, and whose product is 9 ; then the only nu- 
 merator above unity which is left, is 55 ; and, consequent- 
 ly, the whole is brought to the single improper fraction -% 5 -, 
 which is the answer required. And thus let every simi- 
 lar question be managed. 
 
 2. In what time will $350 gain $26^ at 5 per cent. 
 per annum ? Ans. 18 months. 
 
 NOTE.- For Double Rule of Three, Fellowship, &e. 
 in decimals, let the scholar take, as exercises, any such 
 questions in these rules, in whole numbers, as have frac- 
 tions or compound numbers in them. 
 
 SIMPLE INTEREST BY DECIMALS. 
 A TABLE OF RATIOS. 
 
 Rate 
 
 per cent. 
 
 Ratio. | Rate per cent. 
 
 Ratio. 
 
 
 3 
 
 ,03 
 
 4* 
 
 ,055 
 
 
 4 
 
 ,04 
 
 6 
 
 ,06 
 
 
 *i 
 
 ,045 
 
 6j- 
 
 ,065 
 
 
 5 
 
 ,05 
 
 7 
 
 ,07 
 
 Ratio is the simple interest of 1 for 1 year, or, in 
 Federal Money, of $1 for one year, at the rate percent, 
 agreed on. 
 
 RULE. Multiply the Principal, Ratio, and Time con- 
 tinually together, and the last product will be the interest 
 required* 
 
164 SIMPLE INTEREST BY DECIMALS. 
 
 EXAMPLES. 
 
 I. Required the interest of $425,65cts. for 6 years, at 
 6 per cent, per annum. 
 
 $. cts. 
 
 425,65 Principal. 
 ,06 Ratio. 
 
 25,53 90 Interest for one year. 
 6 Multiply by the time. 
 
 153,23 40 Ans.=$153,23cts. 4m. 
 
 2. What is the interest of .645 10s. for 3 years, at 6 
 per cent, per annum ? 
 
 645,5x,06x3=<l 16,190=^116 3s. 9d..2,4qrs. Ans. 
 
 3. Required the amount of $648,50cts. for 12J years, 
 at 5=5- per cent, per annum. Ans. $1103,26cts.+ 
 
 4. What is the amount of $243,39cts. for 1^ year at 6 
 per cent, per annum ? Ans. 8270,7451. 
 
 CASE 2. The amount, time, and ratio given, to Jind 
 the principal. 
 
 RULE. Multiply the ratio hy the time ; add unity to 
 the product for a divisor, by which sum divide the amount, 
 and the quotient will be the principal. 
 
 EXAMPLES. 
 
 1. What principal will amount to $1235,97cts. 5m. in 
 
 5 years, at 6 per cent, per annum 1 
 
 __ <s> <& 
 
 ,06 x 5 + 1 = 1 ,30) 1235,975(950,75 Ans. 
 
 2. What principal will amount to 956 10s. 4,125d. 
 in 8f years, at 5^ per cent. 1 Ans. ,645 15s. 
 
 3. What principal will amount to $l3S4,50cts. in 7 
 years, at 6 per cent, per annum ? Ans. $975. 
 
 CASE 3. The amount, principal, and time given, to 
 Jind the ratio. 
 
 RULE. Subtract the principal from the amount; divide 
 the remainder by the product of the time and principal, 
 and the quotient will be the ratio. 
 
SIMPLE INTEREST BY DECIMALS. 165 
 
 EXAMPLES. 
 
 1. At what rate per cent, will $950,75 amount to 
 $1235,975 in 5 years ? 
 
 From the amount =1235,975 
 Take the principal =950,75 
 
 950,75x5=4753,75)285,2250(,06=6 per cent. Ans. 
 
 2. At what rate per cent, will 543 amount to 705 
 18s. in 5 years ? Ans. 6 per cent. 
 
 3. At what rate per cent, will $2124,25 amount to 
 $3482,44234375 in 7f years ? Ans. 8 per cent. 
 
 CASE 4. The amount, principal, and rate per cent. 
 
 given, .to Jind the time. 
 
 RULE.- Subtract the principal from the amount ; di- 
 vide the remainder by the product of the ratio and prin- 
 cipal ; and the quotient will be the time. 
 
 EXAMPLES. 
 
 1. In what time will $248,39 amount to $270,7451 at 
 6 per cent, per annum 1 
 
 From the amount $270,7451 
 Take the principal 248,39 
 
 248,39x,06= 14,9034)22,355 1(1, 5 = ! year. Ans. 
 
 2. In what time will c543 amount to "^705 18s. at 6 
 per cent, per annum ? Ans. 5 years. 
 
 '3. In what time will $2142,25 amount to $3482,4423- 
 4375 at 8J per cent, per annum 1 Ans. 7f years. 
 
 TO CALCULATE INTEREST FOR DAYS. 
 
 A TABLE OF RATIOS FOR DAYS. 
 
 Rate per cent. R.atios. 
 4^=,0001232S767 
 5~=, 000 1 369863 
 5=,00015068493 
 
 Rate per cent. Ratios. 
 6 =,00016438356 
 6i=,00017808219 
 7" =,00019178082 
 
 Rule. Multiply the principal by the given number of 
 days and that product by the ratio for a year ; divide the 
 last product by 365, (the number of days in a year,) and it 
 
166 SIMPLE INTEREST BY DECIMALS. 
 
 will give the interest required. Or, multiply the ratio 
 for a day in the foregoing table by the principal, and that 
 product by the given number of days ; and the last pro- 
 duct will be the interest required. 
 
 EXAMPLES. 
 
 1. What is the interest of ^300, 10s. for 146 days, at 
 6 per cent. ? 
 
 360,5xl46x,06 . . s. d. qr. 
 --- =8,652=8 13 1,92 Ans. 
 
 365 
 Or, 00016438356x360,5x146 ==8,6519999+Ans. 
 
 2. What is the interest of g780,40cts. for 100 days at 
 6 per cent, per annum ? Ans. $12,S2cts. 8m. + 
 
 3. What is the interest of S48l,75cts. for 25 days at 7 
 percent, per annum ? Ans. $2,30cts. 9m. + 
 
 NOTE. The interest of any sum for 6 days, at 6 per 
 cent, is just as many mills and decimals of a mill, as the 
 principal contains dollars and decimals of a dollar. 
 Therefore set down the principal, multiply it by the days, 
 and divide the product by 6 ; the quotient will be the in- 
 terest in mills and decimals of a mill. This is calling 
 only 30 days a month. 
 
 What is'the interest of g231,84 for 100 days ? 
 231, 84x1 00 days? 
 
 
 6 Or $3,86 4 ; which 
 
 is 5cts. 1m. too much ; but when the time is less than 30 
 days, it gives the answer very exact, for ordinary sums. 
 
 When interest is to be calculated on cash account-:, 
 &c. where partial payments are made, it is the common 
 practice to multiply the several balances into the days 
 they are at interest ; then to multiply the sum of these 
 products by the rate on the dollar, and divide the last 
 product by 365 ; and thus cast the whole interest due on 
 the account, &c. 
 
 EXAMPLE. 
 
 Lent John Joy, per bill on demand, dated 1st of June, 
 1821, $2000, of which I received back the 19th of August, 
 $400 ; on the 15th of October, $600; on the llth of De- 
 cember, 400 ; on the l7th of February, 1822, $200 ; 
 and on the 1st of June, $400 ; how much interest is due 
 on the bill, reckoning at 6 per cent. ? 
 
COMPOUND INTEREST BY DECIMALS. 
 
 1821. 
 
 June 1. Principal, per bill, 
 Aug. 19. Received in part, 
 
 Balance, 
 Oct. 15. Received in part 
 
 Balance, 
 Dec. 11. Received in part, 
 
 1822, Balance, 
 
 Feb. 17. Received in part, 
 
 Balance, 
 
 June 1. Rec'd in full of principal, 400 
 Then 388600 
 
 ,06 Ratio. 
 
 $ cts.m. 
 
 365)23316,00(63,879+ Ans. =63,87 9 
 
 167 
 
 $ 
 
 2000 
 400 
 
 days. 
 79 
 
 57 
 
 products. 
 158000 
 
 91200 
 
 1600 
 600 
 
 1000 
 400 
 
 57 
 
 57000 
 
 600 
 200 
 
 68 
 
 40800 
 
 400 
 , 400 
 
 104 
 
 41600 
 
 388600 
 
 COMPOUND INTEREST BY DECIMALS. 
 
 A table showing the amount of 1 or $1 at 5 and 6 per 
 
 cent, per annum, compound interest, for 20 years. 
 
 Yrs. 
 
 5 per cent. 
 
 6 per ct. 
 
 Yrs. 
 
 5 per cent. 
 
 6 per cent. 
 
 1 
 
 ,05000 
 
 1,06000 
 
 11 
 
 1,71033 
 
 1,89829 
 
 2 
 
 ,15250 
 
 1,12360 
 
 12 
 
 1,79585 
 
 2,01219 
 
 3 
 
 ,15762 
 
 1,19101 
 
 13 
 
 1,88564 
 
 2,13292 
 
 4 
 
 ,21550 
 
 1,26247 
 
 14 
 
 1,97993 
 
 2,26090 
 
 5 
 
 ,27628 
 
 1,33822 
 
 15 
 
 2,07892 
 
 2,39^55 
 
 6 
 
 ,34009 
 
 1,41851 
 
 16 
 
 2,18287 
 
 2,54035 
 
 7 
 
 ,40710 
 
 1,50363 
 
 17 
 
 2,29201 
 
 2,69277 
 
 8 
 
 ,47745 
 
 1,59384 
 
 18 
 
 2,40661 
 
 2,85433 
 
 9 
 
 ,55132 
 
 1,68947 
 
 19 
 
 2,52695 
 
 3,02559 
 
 10 
 
 1,62889 
 
 1,79084 j 20 
 
 2,65329 
 
 3,20713 
 
 RULE. Multiply the given principal continually by the 
 amount of one pound, or one dollar, for one year, at the 
 rate per cent, given, until the number of multiplications 
 is equal to the given number of years, and the product 
 will be the amount required. 
 
168 ANNUITIES AT COMPOUND INTEREST. 
 
 Or, Take from the preceding Table, the amount of one 
 pound, or one dollar, as the case may be, for the given 
 number of years, and at the given rate per cent., and 
 multiply it by the given principal, and it will give the 
 amount as before. 
 
 EXAMPLES. 
 
 1. What will 400 amount to in 4 years, at 6 per cent, 
 per annum, compound interest? 
 
 400xl,06xl,06xl,06xl,06=504,99+ or ,504 19s. 
 
 9d. 2,4qrs.-f Ans. 
 Or, by the Table. Tabular amount of ,1 = 1,26247 
 
 Multiply by the principal 400 
 
 Whole amount 504,98800 
 
 2. What is the compound interest of 555 for 14 years, 
 at 5 per cent. ? Ans. 543,S6cts.+ 
 
 NOTE. Any sum of money, at 6 per cent, per annum, 
 simple interest, will double in 16f years; but at 6 per 
 cent, per annum compound interest, it will double in 11 
 years and 325 days, or 11,889 years. 
 
 ANNUITIES AT COMPOUND INTEREST. 
 
 CASE 1. To find the, amount of an annuity, fyc. 
 
 RULE. Raise the amount of $1, or JBl, at the given 
 rate per cent., for one year, to that poicer denoted by the 
 given number of years ; subtract unity or 1 from this pro- 
 duct ; multiply the remainder by the given annuity ; di- 
 vide this last product by the ratio made less by unity or 
 1 ; and the quotient will be the amount sought. 
 
 EXAMPLES. 
 
 1. If $250, yearly pension, be foreborne 7 years, what 
 will it amount to, at 6 per cent, per annum compound in- 
 terest ? 
 
 1,06x1,06x1,06x1,06x1,06x1,06x1,06 I,x250 
 
 1,061, 
 
 $2098,45c. 9m. -f Ans. 
 
INVOLUTION. 169 
 
 2. If a salary, or an annuity, .of ^100 per annum, runs 
 on unpaid for 6 years, at 5 per cent, compound interest, 
 what is the amount due at the end of that period ? 
 
 Ans. (iSO 3s. 9d. ,63. 
 
 CASE 2. To find the present worth of an Annuity, fye. 
 R ULE . Raise the amount of gl, or l, at the given 
 rate per cent., for 1 year, to that power denoted by the 
 given number of years ; divide the given annuity by this 
 product; subtract its quotient from the given annuity; 
 divide the remainder by the ratio made less by unity or 
 1 ; and the quotient will be the present worth sought. 
 EXAMPLES. 
 
 1. What is the present worth of a salary of $300, to 
 continue 5 years, at 5 per cent, compound interest ? 
 
 300 
 
 ---- =235,0578499405. + 
 1,05x1,05x1,05x1,05x1,05 
 
 300235,0578499405 g c. m. 
 Then, --- =1298,84 3+Ans. 
 1,051, 
 
 2. What is the present worth of ^630 per annum, to 
 continue 7 years, at 6 per cent, compound interest 1 
 
 Ans. ^'167 9s. 5d.+ 
 
 INVOLUTION. 
 
 INVOLUTION is the continual multiplication of a num- 
 ber into itself; and the products thence arising, with the 
 original number itself, are called the powers of that num- 
 ber. 
 
 Any number may itself be called a first power. If the 
 first power be multiplied by itself, the product is called 
 the second power, or square ; if the square be multiplied 
 by the first power, the product is called the third power, 
 or cube ; if the cube be multiplied by the first power, the 
 product is called the fourth power, or biquadrate, &,e. 
 
 Thus 3 is the first power of 3. 
 3x3=9 is the second power of 3. 
 3x3x3=27 is the third power of 3. 
 3x3x3x3=81 is the fourth power of 3 &c. <fcc. 
 And in this manner is formed the folio wing table of powers. 
 
170 EVOLUTION, OR EXTRACTION OF ROOTS. 
 
 Table of the SQ UARES and CUBES of tlie nine digits. 
 
 Roots. 
 Squares. 
 Cubes. 
 
 1 
 
 2 
 4 
 
 8 
 
 3 
 9 
 
 27 
 
 4 
 16 
 64 
 
 5 
 
 6 
 
 7 
 
 8 
 
 9 
 
 
 1 
 1 
 
 25 
 
 36 
 
 49 
 
 64 
 
 81 
 
 125 
 
 216 
 
 343 
 
 512 
 
 729 
 
 EXAMPLES. 
 
 1. What is the 6th power of 8? 
 8 the root, or 1st power. 
 
 8 
 
 64 = 
 
 8 
 
 =2d power, or square. 
 
 NOTE. The num- 
 ber denoting the 
 height of the power 
 is called the index, 
 or exponent of that 
 power; so the 2d 
 power of 3 may be 
 denoted by 3 2 , the 3d 
 by 3 3 , the 4th by 3 4 , 
 
 &c. ; the 6th power 
 
 4096 =4th power, or biquadrate. of 8 by8 6 , &c. 
 8 
 
 5l2=3d power, or cube. 
 
 8 
 
 32768 =5th power, or sursolid. 
 8 
 
 262144=6th power, or square cube. Ans. 8 6 . 
 
 2. What is the 7th power of f ? Ans. 
 
 3. What is the 5th power of | or l ? Ans. 
 
 4. What is the fourth power of ,27 ? Ans. ,00531441. 
 
 EVOLUTION, OR EXTRACTION OF ROOTS. 
 
 WHEN the root of any power is required, the business 
 of finding it is called the extraction of the Root. 
 
 The root is that number, which by a continual multi- 
 plication into itself, produces the power which is given to 
 be extracted. 
 
 %Though every number will produce a perfect power by 
 involution, yet there are many numbers, the precise roots 
 of which can never be determined. By the help of deci- 
 mals, however, we can approximate towards the root, to 
 any assigned degree of exactness. 
 
EXTRACTION OF THE SQUARE ROOT. 171 
 
 The roots which approximate, are called surd roots, 
 and those which are perfectly accurate, are called rational 
 roots, 
 
 TO EXTRACT THE SQUARE ROOT. 
 
 Any number multiplied into itself, produces a square. 
 The extracting of the square root, is only finding a num- 
 ber, which, being multiplied into itself, shall produce the 
 given number. 
 
 RULE. 1. Distinguish the given number into periods 
 of two figures each, by putting a point over the place of 
 units, another over the place of hundreds, and so on over 
 every second figure ; and if there be decimals, point them 
 in the same manner, from units towards the right hand ; 
 which points show the number of figures the root will con- 
 sist of. 
 
 2. Find by the table or trial the greatest square num- 
 ber in the first, or left hand period, place the root of it at 
 the right hand of the given number, (after the manner of 
 a quotient in division,) for the first figure of the root ; 
 and set the square number under the period, subtract it 
 therefrom, and to the remainder bring down the next pe- 
 riod for a dividend. 
 
 3. Place the double of the root already found on the 
 left hand of the dividend for a divisor. 
 
 4. Consider what figure must be annexed to the divis- 
 or, so that if the result be multiplied by it the product 
 may be equal to, or the next less than, the dividend, and 
 it will be the second figure of the root. 
 
 5. Subtract that product from the dividend, and to the 
 remainder bring down the next period for a new dividend. 
 
 6. Find a divisor as before, by doubling the figures al- 
 ready in the root ; and from these find the next figure of 
 the root as in the last article ; and so on through all the 
 periods to the last. 
 
 Or, to facilitate the foregoing operation, when a period 
 is brought down to a remainder, and a dividend thus form- 
 ed, in order to find a new figure in the root, divide said 
 dividend, (omitting the right hand figure thereof,) by do*u- 
 ble the root already found, and the quotient will common- 
 ly be the figure of the root sought, or, being made less by 
 one, or two, will generally give the next figure sought, 
 
172 EXTRACTION OF THE SQUARE ROOT. 
 
 To extract the square root of a Vulgar Fraction. 
 First prepare all vulgar fractions by reducing them to 
 their lowest terms, both for this and all other roots. Then, 
 
 1. Take the root of the numerator and that of the de- 
 nominator for the respective terms of the root required ; 
 and this is the best way if the denominator be a complete 
 power. But if not, 
 
 2. Multiply the numerator and denominator together ; 
 take the root of the product ; this root, being made the 
 numerator to the denominator of the given fraction, or 
 the denominator to the numerator of it, will form the 
 fractional root required. 
 
 3. Or raduce the vulgar fraction to a decimal, and ex- 
 
 tract its root. 
 
 EXAMPLES. 
 1. Required the square root of 6749604. 
 
 6749604(2598 Ans. The root exactly without a 
 4 remainder ; but when the periods be- 
 
 - - longing to any given number are all 
 
 45)274 exhausted, arid still leave a remain- 
 
 5)225 der, the operation may be continued 
 
 --- at pleasure, by annexing periods of 
 
 509)4996 ciphers, &c. Roots are often denot- 
 
 9)4-581 ed by writing v/ before the power, 
 
 with the index of the root within or 
 5188)41504 over it, save the index of the square 
 
 8)41504 root, which is ever understood : so v/ 
 
 64 is the 2d or square root of 64 ; 
 
 2, Required the 64 the 3d or cube root of 64 ;> </ 64 
 square root of 739,4 the 4th root of 64. 
 
 739,40(27,19 + root. . 
 
 4 When the square root of a number 
 
 is wanted to many places, the work 
 47)339 may be much abridged. Find half 
 
 7)329 the root by the rule ; then to get the 
 
 rest, annex to the last remainder as 
 
 541) 1040 many ciphers as you need, and divide 
 
 1) 541 it by the double of the root before 
 
 found. 
 
 5429)49900 
 9)48861 
 
 1039 remainder. 
 
APPLICATION AND USE OF THE SQUARE ROOT. 173 
 
 3. What is the square root of 2 ? Ans. 1,41421356.+ 
 
 4. What is the square root of 10342056? Ans. 3216. 
 
 5. What is v/964,5192360241 ? 
 
 Ans. 31,05671. 
 
 6. What is v/,00032754? 
 
 Ans. ,01809.+ 
 
 7. What is \/fHt- Ans - rV 
 
 8. What is x/42J ? Ans. 8f 
 
 9. What is x/6f 1 Ans.2,5298+&c. 
 
 APPLICATION AND USE OF THE SQUARE 
 ROOT. 
 
 CASE 1. To Jind a mean proportional between any two 
 
 given numbers. 
 
 RULE. Multiply the two given numbers together, and 
 extract the square root of the product, which root will 
 be the number sought. 
 
 EXAMPLE. 
 What is the mean proportional between 16 and 36 ! 
 
 v/36x!6=24. Ans. 
 
 CASE 2. To find the side of a square equal in area to 
 any given superficies. 
 
 RULE. Extract the square root of the given superficies* 
 which root will be the side of the square sought. 
 
 EXAMPLES. 
 
 1. If an acre of land contains 160 square rods, what 
 will be the side of a square, which should contain just aa 
 acre ? v/ 160 = 12,649 + rods. Ans. 
 
 2. A general having an army of 5184 men wishes to 
 form them into a square ; how many must he place ia 
 rank and file ! v/5 1 84 =72. Ans. 
 
 3. Let 8192 men be formed into an oblong, so that the 
 mmvber in rank may be double the file. 
 
 8192 
 
 -v/ - =64 in fite. 64x2=123 io rank. 
 2 
 P2 
 
174 APPLICATION AND USE OF THE SQUARE ROOT. 
 
 4. Suppose a gentleman would set out an orchard of 
 864 trees, so that the length shall be to the breadth as 3 
 to 2, and the distance of each tree, one from the other, 7 
 yards ; how many trees must there be in length, and how 
 many in breadth, and how many square yards of ground 
 do they stand on 1 
 
 To resolve any question of this nature, say, as the ra- 
 tio in length is to the ratio in breadth, so is the number of 
 trees to a fourth number, whose square root is the number 
 in breadth ; then as the ratio in breadth is to the ratio in 
 length, so is the number of trees to a fourth number, 
 whose root is the number in length. And as unity is to 
 the distance, so is the number in length less by one to a 
 fourth number ; next do the same by the breadth, and 
 multiply the two numbers thus found together, and the 
 product will be the answer. 
 
 As 3 : 2 : : 864 : 576,& ^7576=24 num. in breadth. Ans. 
 As 2 : 3 : : 864 : 1 296, &V 1 296 =36 num. in length Ans. 
 As 1 : 7 : : 361 : 245, And, as 1 : 7 : : 241 : 101. 
 And 245x161=39445 square yards. Ans. 
 
 CASE 3. To ascertain the proportionate capacities of 
 
 water pipes. 
 
 RULE. Square the given diameter, and multiply said 
 square by the given proportion ; the square root of the 
 product is the answer. 
 
 EXAMPLE. 
 
 Admit lOhhds. of water are discharged through a leaden 
 pipe of 2J- inches diameter, in a certain time ; what must 
 be the diameter of another pipe, that shall discharge four 
 times as much water in the same time ? 
 
 2!=2,5 and 2,&x 2,5 = 6,25 square. 
 
 4 given proportion. 
 
 v/25,00=5 inches diam. Ans. 
 
 CASE 4. The sum of any two numbers, and their product 
 
 being given , to find each number. 
 
 RULE. From the square of their sum, subtract 4 times 
 their product, and extract the square root of the remain- 
 
APPLICATION AND USE 07 THE SQUARE ROOT. 175 
 
 der, which will be the difference of the two numbers ; 
 then half the said difference added to half the sum, gives 
 the greater of the two numbers, and the said half differ- 
 ence, subtracted from the half sum, gives the less number. 
 
 EXAMPLES. 
 
 1. The sum of two numbers is 46, and their product 
 is 504 ; what are those two numbers 1 
 
 The sum of the numbers 46x46=2116 sq. of their sum. 
 The product of ditto. 504x4=2016 four times the pro. 
 
 [numbers. 
 
 ^/100 = 10 differ, of the 
 
 462=23 half sum. 10-f-2=5 half differ. 
 
 +5 half diff. 23 half sum. 
 
 5 half differ. 
 28 greater number. Ans. 
 
 18 less number. Ans. 
 
 2. Bought a certain quantity of broadcloth for $573, 
 75cts. ; and if the number of cents which it cost per 
 yard, was added to the number of yards bought, the sum 
 would be 480 ; how many did I buy, and at what price 
 per yard. Ans. 255yds. at $2,25cts. per yard. 
 
 3. If I lay out a lot of land in an oblong form, con- 
 taining 7 acres, 1 rood, and 10 rods, and taking just 142 
 rods of wall to enclose it ; pray how many rods long, and 
 how many wide is said lot ? 
 
 Ans. 45 rods long, and 26 rods wide. 
 
 CASE 5. To Jind the degree of light, heat, or attraction. 
 
 NOTE. The effects on degrees of light, heat, and attrac- 
 tion, are in proportion* to the squares of the distance, 
 whence they are propagated. 
 
 EXAMPLES. 
 
 1. Two men, A and B, are sitting in a room, the for- 
 mer 3, and the latter 6 feet distant from a fire ; how 
 much hotter is it at A's, than at B's seat ? 
 3x3=9,& 6x6=36. Then, as 9 : 1 : : 36 : 4 r so that 
 
 A's place is 4 times as hot as B's. Ans. 
 
 2. If the earth's mean distance from the sun be 95,000- 
 000 of miles, at what distance from him must another 
 
176 APPLICATION AND USE OF THE SQUARE ROOT. 
 
 body be placed, that it may receive a degree of light and 
 heat, double to that of the earth 1 
 
 95 OOOOOO 2 
 
 \/ =67175144 + mile. Ans. which is somewhat 
 
 2 less than the distance of Venus from the sun. 
 
 3. A ball descending by the force of gravity from the 
 top of a tower, was observed to fall half the way in the 
 last second of time ; how long was it in descending, and 
 what was the height of the tower ? 
 
 The square roots of the distance are as the times, viz. 
 As the 5/1 : -v/2 : : the time of falling through : the whole 
 required height. 
 
 Now, the x/l = l, and ^2 = 1,4142, from which take 
 1 ; ,4142 remains. 
 
 And, as ,4142 : 1,4142 : : 1 : 3,414 + sec. time of de- 
 scent ; the square of which is 11,6554 nearly. And the 
 velocity acquired by heavy bodies falling near the surface 
 of the earth, is 16 feet in the first second, 64 in the second 
 second, 144 in the third second, &/c. that is, the space 
 fallen through [in feet] is always equal to the square of 
 the time in 4ths of a second, 
 ft. sec. sq. 
 
 As I 2 : 16 : : 11,6554 : 186,4864 = 186^ feet nearly, 
 height of the tower, Ans. 
 
 CASE 6. Any two sides of a right-angle triangle given 
 to find the other side. 
 
 RULE. Extract the square root of the sum of the 
 squares of the two least sides, and that root is the great- 
 est side ; for the square root of the sum of the squares of 
 the two legs, is always the length of the hypotenuse. 
 Extract the square root of the difference of the squares 
 of either of the two least sides and the greatest side, and 
 that root is the other side ; for the square root of the dif- 
 ference of the squares of either leg and the hypotenuse, is 
 always the length of the other leg. 
 
 EXAMPLES. 
 
 1. A ladder 40 feet long may be so planted as to reach 
 a window 33 feet from the ground, on one side of the 
 street; and, without moving it at the foot, will do the 
 same by a window 21 feet high on the other side; how 
 wide is the street ? 
 
 402=xl600. 332=1089. 212=441. Then 16001089 
 
EXTRACTION OF THE CUBE ROOT. 177 
 
 =511, and x/51 1=22,6; and 1600441=1159, and 
 v/1 159=34,04; then 22,6+34,04=54,64 f ee t.+ Ans. 
 
 2. A line 27 yards long will exactly reach from the 
 top of a fort to the opposite bank of a river, known to be 
 23 yards broad ; what is the height of the wall ? 
 
 Ans. 14,142+yards. 
 
 3. Two ships sail from the same port ; one sails due 
 east 50 leagues, and the other due north 84 leagues ; how 
 far are they then apart ? Ans. 97,75+ leagues. 
 
 TO EXTRACT THE CUBE ROOT. 
 
 A cube is any number multiplied by its square. 
 
 To extract the Cube Root, is to find a number which, 
 being multiplied into its square, shall produce the giveji 
 number. 
 
 RULE. 1. Separate the given number into periods of 
 three figures each, by putting a point over the unit figure, 
 and every third figure both ways from the place of units. 
 
 2. Find the nearest less cube to the first period by the 
 table of powers or trial ; set its root in the quotient ; sub- 
 tract the cube found from the first period, and to the re- 
 mainder bring down the second period, and call this the 
 resolvend. 
 
 3. To three times the square of the root just found, 
 add three times the root itself, setting this one place more 
 to the right than the former, and call this sum the divisor. 
 Then divide the resolvend, omitting the unit figure, by 
 the divisor, for the next figure of the root, which annex 
 to the former, calling this last figure c, and the part of 
 the root before found call a. 
 
 4. Add together these three products, viz. thrice the 
 square of a multiplied by e, thrice a multiplied by the 
 square of e, and the cube of e, setting each of them one 
 place more to the right hand than the former, and call 
 the sum the subtrahend, which must not exceed the resol- 
 vend ; but if it do, then make the last figure e less, and 
 repeat the operation for finding the subtrahend. 
 
 5. From the resolvend take the subtrahend, and to the 
 remainder join the next period of the given number for a 
 
178 APPLICATION AND USE OF THE CUBE ROOT. 
 
 new resolvend ; to which form a new divisor from the 
 whole root now found, and thence another figure of the 
 root as before, &,c. 
 
 EXAMPLES. 
 1. Required the cube root of 48228,544. 
 
 3x32=27 
 3x3 = 09 
 
 Divisor 279 
 
 48228,544(36,4 root. 
 
 27 
 
 21228 resolvend. 
 
 3x3 2 x6=162 ) 
 3x3x6 2 = 324 VAdd. 
 6*= 216 j 
 
 3x362=3888 
 3x36 = 108 
 
 - 
 
 19656 subtrahend. 
 
 Divisor 38988 
 
 1572544 resolvend. 
 
 3x36 2 x4 = 15552 V 
 3x36 x42 1728 } Add. 
 
 4 3 =5 64 J 
 
 1572544 subtrahend. 
 
 2. What is the cube root of 1092727? Ans. 103. 
 
 3. What is ^34965783 ? Ans. 327. 
 
 4. What is ^/,0001357 ? 
 
 Ans. ,05138.+ 
 
 5. What is -NXfriny Ans - 
 
 6. What is ^/f ? Ans. ,873.+ 
 
 7. What is x 3/-2? Ans. 1,25.+ 
 
 8. What is 3'*i* ? Ans. f. 
 
 APPLICATION AND USE OF THE CUBE ROOT. 
 
 EXAMPLES. 
 
 1. The statute bushel contains 2150,4197+ cubic or 
 solid inches ; I demand the side of a cubic box which 
 shall contain just that quantity ? 
 
 ^/2150,419724=12,907+inch. Ans. 
 
 NOTE. The solid contents of similar figures, are in 
 proportion to each other, as the cubes of their similar 
 sides or diameters. 
 
TO EXTRACT THE ROOTS OF POWERS IN GENERAL. 179 
 
 2. If a bullet 4 inches diameter weigh 9^., what will a 
 bullet of the same metal weigh,whose diameter is 8 inches? 
 4x4x4=64 Sx8x8=5l2. As 64 : 9 : : 512 : 72ft. Ans. 
 
 3. If a solid globe of silver, of 3 inches diameter, be 
 worth $150, what is the value of another globe of silver, 
 whose diameter is eight inches ? 
 
 3x3x3=27. 8x8x8=512. As 27 : 8150 : : 512 : 
 
 $2844J-f Ans. 
 
 The side of a cube being given, to find the side of that 
 cube which shall be double, triple, &c. in quantity to the 
 given cube. 
 
 RULE. Cube the given side, and multiply it by the 
 given proportion between the given and required cube, 
 and the cube root of the product will be the side sought. 
 
 4. If a cube of silver, whose side is 2 inches, be worth 
 820, what should the side of a cube of like silver be, 
 whose value would be 8 times as much ? 
 
 2x2x2=8, and 8x8=64. ^64=4 inches. Ans. 
 
 5. There is a cubical vessel whose side is 4 feet ; I de- 
 mand the side of another cubical vessel, which shall con- 
 tain 4 times as much 1 
 
 4x4x4=64. & 64x4=256. -^256 =6,349+ feet. Ans. 
 
 6. A cooper having a cask 40 inches long, and 32 in- 
 ches at the bung diameter, is ordered to make another 
 cask of the same shape, but which shall hold just twice 
 as much ; what will be the bung diameter, and length of 
 the new cask 1 
 
 40x40x40x2=128000; then ^/ 128000 =50,3+ in- 
 ches length. Ans. 
 
 32x32x32x2=65536; <fc ^65536=40,3 + inches 
 
 bung diam. Ans. 
 
 TO EXTRACT THE ROOTS OF POWERS IN GENERAL. 
 
 RULE. 1. Prepare the given number for extraction by 
 pointing oft" from the place of units as the root required 
 directs ; 4th root put a dot over every 4th figure &,c. from 
 the place of units ; 5th root, over every 5th, &c. from 
 units' place, &c. 
 
 2. Find the first figure of the root by trial, and sub- 
 tract its power from the given number. 
 
180 TO EXTRACT THE ROOTS OF POWERS IN GENERAL. 
 
 3. To the remainder bring down the first figure in the 
 next period, and call it the dividend. 
 
 4. Involve the root to the next inferiour power to that 
 which is given, and multiply it by the number denoting 
 the given power* for a divisor. 
 
 5. Find how many times the divisor maybe had in the 
 dividend, and the quotient will be another figure of the root. 
 
 6. Involve the whole root to the given power, and sub- 
 tract it from the given number as before. 
 
 7. Bring down the first figure of the next period to the 
 remainder for a new dividend, to which find a new divi- 
 sor, and so on, until the whole is finished. 
 
 EXAMPLES. 
 1. What is the cube root of 53157376 ? 
 
 53 157376(376 root. 
 27=3 3 
 
 3 2 x3=27)26i dividend. 
 50653 =37 3 
 
 37 2 x3=4107)25043 second dividend. 
 
 53157376=376 3 
 
 2. What is the biquadrate root of 34827998976 ? 
 
 Ans. 431,9.+ 
 
 3. What is the sursolid root of 281950621875 ? 
 
 Ans. 195. 
 
 4. What is the square cubed, or sixth root of 16196, 
 005304479729 ? Ans. 5,03. 
 
 5. Find the seventh root of 34487717407,30751. 
 
 Ans. 32,0 i.+ 
 
 NOTE. The roots of most powers may be found by the 
 square and cube roots only ; therefore, when any even 
 power is given, the better way will be,~ especially in very 
 high powers, to extract the square root of it, which re- 
 duces it to half the given power, then the square root of 
 that power ; and so on till it comes to a square or cube. 
 
 For example, suppose a 12th power be given ; the 
 square root of that reduces it to a sixth power ; and the 
 square root of a sixth power to a cube. 
 
ARITHMETICAL PROGRESSION 181 
 
 6. Extract the eighth root of 7213895789838336. 
 
 Ans. 96. 
 
 7. What is the biquadrate root of 5308416 ? Ans. 48. 
 
 ARITHMETICAL PROGRESSION. 
 
 ANY rank of numbers more than two, increasing by a 
 common excess, or decreasing by a common difference, is 
 said to be in Arithmetical Progression : such are the 
 numbers 1, 2, 3, 4, &c. 7, 5, 3, 1 ; and ,8, ,6, ,4, ,2. 
 When the numbers increase, they form an ascending series ; 
 but when they decrease, they form a descending series. 
 
 The numbers which form the series, are called the 
 terms of the progression. 
 
 Any three of the five following terms being given, the 
 other two may readily be found. 
 
 1st. The first term, ) 
 
 3d. The last term, } commonl 7 -C a ed the extremes. 
 
 3d. The number of terms. 
 4th. The common difference. 
 5th. The sum of all the terms. 
 
 PROBLEM 1. 
 
 The first term, the last term, and the number of terms be- 
 ing given, to Jind the sum of all the terms. 
 RULE. Multiply the sum of the extremes by the num- 
 ber of terms, and half the product will be the answer. 
 
 EXAMPLES. 
 
 1. The first term of an arithmetical progression is 1, 
 the last term 21, the number of terms 11 ; required the 
 sum of the series. 
 
 21 
 1 
 
 22 
 
 11 21 + lxll 
 
 Or =121 the Ans. 
 
 2)242 2 
 
 Ans. 121 
 Q 
 
182 ARITHMETICAL PROGRESSION. 
 
 2. How many strokes does a Venice clock strike in the 
 compass of a day, going to 24 o'clock ? Ans. 300. 
 
 3. If 100 stones be placed in a right line, a yard dis- 
 tant from each other, and the first a yard from a basket ; 
 what distance will that man go who gathers them up sing- 
 ly, returning with them one by one to the basket 1 
 
 Ans. 5 miles and 1300 yards. 
 
 4. A draper sold 100 yards of cloth at5cts. for the first 
 yard, lOcts. for the second, 15 for the third, &c., increas- 
 ing 5cts. for every yard ; what did the whole amount to, 
 and what did it average per yard ? 
 
 Ans. Amount was $252^, and the average price was 
 
 $2,52cts. 5m. per yard. 
 
 PROBLEM II. 
 
 The extremes and the number of terms being given, to find 
 
 the common difference. 
 
 RULE. Divide the difference of the extremes by the 
 number of terms less 1, and the quotient will be the 
 common difference. 
 
 EXAMPLES. 
 
 1. The extremes are 3 and 19, and the number of 
 terms is 9 ; required the common difference, and the sum 
 of the whole series. 
 
 9 19 19+3x9 
 
 I 3 And =99 the sura. 
 
 2 
 
 8) 16 
 
 2 difference. 
 
 2. A man is to travel from Boston to a certain place in 
 12 days, and to go but three miles the first day, increasing 
 every day by an equal excess, so that the last day's jour- 
 ney may be 58 miles ; required the daily increase, and 
 the distance of the place from Boston. 
 
 Ans. Daily increase 5, distance 366 miles. 
 2. A man had 12 sons whose several ages differed 
 alike ; the eldest was 49, the youngest 5 years old ; what 
 was the common difference of their ages ? Ans. 4 years. 
 
GEOMETRICAL PROGRESSION. 183 
 
 PROBLEM III. 
 Given the first term, the last term, and the common differ- 
 
 ence, to find the number of terms. 
 
 RULE. Divide the difference of the extremes by the 
 common difference, and the quotient increased by 1, is 
 the number of terms required. 
 
 EXAMPLES. 
 
 1. The extremes are 3 and 19, and the common differ- 
 ence 2 ; what is the number of terms ? 
 
 19 
 3 
 
 2)16 193 
 
 Q r -- f. i 9 the Answer. 
 8 2 
 
 1 
 
 Ans. 9 
 
 2. Suppose a man travel the first day 7 miles, the last 
 51 miles, and increase his journey each day by 4 miles ; 
 how many days will he travel, and how far ? 
 
 Ans. 12 days, and 343 miles. 
 
 GEOMETRICAL PROGRESSION. 
 
 ANY series of numbers, the terms of which gradually 
 increase or decrease by a constant multiplication or di- 
 vision, are said to be in Geometrical Progression. Thus, 
 4, 8, 16, 32, 64, &c. and 81, 27, 9, 3, 1, &c. are series 
 in geometrical progression, the one increasing by a con- 
 stant multiplication by 2, and the other decreasing by a 
 constant division by 3. 
 
 The number by which the series is constantly increas- 
 ed or diminished, is called the ratio. 
 
 PROBLEM I. 
 Given the first term, the last term, and the ratio, to find 
 
 the sum of the series. 
 
 RULE. Multiply the last term by the ratio, and from 
 the product subtract the first term, and the remainder di- 
 vided by the ratio less 1, will give the sum of the series. 
 
184 GEOMETRICAL PROGRESSION. 
 
 EXAMPLES. 
 
 1. The extremes of a geometrical progression are 1 and 
 65536, and the ratio 4 ; what is the sum of the series ? 
 65536 
 4 
 
 262144 4x655361 
 
 1 Or =87381 Ans. 
 
 41 
 41=3)262143 
 
 87381 Ans. 
 
 2. A man travelled 6 days ; the first day he went 4 
 miles, and doubling his travel each day, his last day's ride 
 was 128 miles ; how far did he go in the whole ? 
 
 Ans. 252 miles. 
 
 3. The extremes of a geometrical series are 1024 and 
 59049, and the ratio is l ; what is the sum of the series 1 
 
 Ans. 175099. 
 
 PROBLEM II. 
 
 Given the first term and the ratio, to find any other term 
 assigned.* 
 
 CASE 1. When the first term of the series, and the ratio, 
 are equal.^ 
 
 RULE. 1. Write down a few of the leading terms of 
 the series, and place their indices over them, beginning 
 the indices with a unit or 1. 
 
 2. Add together such indices as, in their sum, shall 
 make up the entire index to the term required. 
 
 * As the last term in a long 1 series of numbers is very tedious to 
 be found by continual multiplication, it will be necessary for 
 more readily finding- it out, to have a series of numbers in arith- 
 metical proportion, called indices, whose common difference is 1. 
 
 f When the first term of the series, and the ratio, are equal, the 
 indices must beg-in with a unit, and in this case, the product of 
 any two terms is equal to that term, signified by the sum of their 
 indices. 
 
 rp, ( 1, 2, 3, 4, 5, &c. Indices arithmetical series. 
 : \ 2, 4, 8, 16, 32, &c. Geometrical series. 
 
 Now, 2+3=5, the index of the fifth term, and 4x8= 
 32=the fifth term. 
 
GEOMETRICAL PROGRESSION. 183 
 
 3. Multiply together the terms of the geometrical se- 
 ries belonging to those indices, and the product will be 
 the term required. 
 
 EXAMPLES. 
 
 1. The first term of a geometrical series is 2, and the 
 ratio 2; required the 13th term. 
 
 1, 2, 3, 4, 5, 6, 7, indices. 
 
 2, 4, 8, 16, 32, 64, 128, leading terms. 
 
 Then 6+7=index to the 13th term. 
 And 64x123=8192 the answer. 
 
 2. A young man agreed with a farmer to work for him 
 11 years, with no other reward than the produce of one 
 grain of wheat for the first year, allowing the increase to 
 he tenfold, and that produce to be sowed the second year, 
 and so on from year to year, until the end of the time ; 
 what is the sum of the whole produce, allowing 7680 
 grains to make a pint, and what does it amount to at one 
 dollar and fifty cents per bushel 1 
 
 Ans. 2^6056i- + busli., and $339084, 19cts.+ 
 
 NOTE. In such questions, you first find the last term 
 by one of the cases in Problem 2, and then the sum of 
 the whole series by Problem 1. 
 
 3. A rich miser thought 20 guineas apiece too much 
 for 12 fine horses, but readily agreed to give 4 cents for 
 the first, 16 cents for the second, 64 cents for the third 
 horse, and so on, in fourfold proportion, to the last ; 
 what did they come to at that rate, and how much did 
 they cost per head, one with another ? 
 
 The \> horses came to S223f>96,20cts., and 
 the average price was 18641,35cts. per head. 
 
 CASE 2* When the jirst term of the series , and the ra- 
 tio > are different: that is, when the first term is either 
 greater or less than the ratio* 
 
 RULE. 1. Write down a few of the leading terms of 
 the series, aa before, and begin their indices with a ci- 
 pher ; thus: 0, 1, 2, 3, &c. 
 
 * When the first term of the series, and the ratio are different, 
 the indices must beg-in with a cipher, and the sum of the indices 
 made choice of, must be one less than the number of terms given 
 in the question ; because 1 in the indices stands orer the second 
 term, and 2 ui the indices over the third term, &c. j and. ia this; 
 
 Q2 
 
186 GEOMETRICAL PROGRESSION. 
 
 2. Add together the most convenient indices to make 
 an index, less by 1, than the number expressing the place 
 of the term sought. 
 
 3. Multiply the terms of the geometrical series togeth- 
 er, belonging to those indices, and make the product a 
 dividend. 
 
 4. Raise the first term to a power whose index is one 
 less than the number of terms multiplied, and make the 
 result a divisor. 
 
 5. Divide the said dividend by the said divisor, and 
 the quotient is the term required. 
 
 NOTE. If the first term of any series be unity, or 1, 
 the term required is found by multiplying the terms of 
 the geometrical series together which belong to those in- 
 dices, without neecfing any division. 
 
 EXAMPLES. 
 
 1. Required the 12th term of a geometrical series, 
 whose first term is 3, and ratio 2. 
 
 0, 1, 2, 3, 4, 5, 0, indices. 
 
 3, 6, 12, 24, 48, 96, 192, leading terms. 
 
 Then, 6+5=index to the 12th term. 
 
 And 192x96 = l8432*=dividend. 
 
 The number of terms multiplied is 2, and 2 1=1 is 
 the power to which the term 3 is to be raised ; but the 
 first power of 3 is 3=divisor ; therefore 
 
 184323=6144, the 12th term. 
 
 2. A goldsmith sold Ijfo. of gold, at 2cts. for the first 
 ounce, Sets, for the second, 32cts. for the third, &c., in 
 quadruple proportion geometrical ; what did the whole 
 come to 1 Ans. $111848,10cts. 
 
 3. A man bought a horse, and by agreement was to give 
 a farthing for the first nail, two for the second, four for 
 the third, &c. There were four shoes, and eight nails 
 in each shoe ; What did the horse come to at that rate ? 
 
 Ans. 4473924 5s. 3d. 3qrs. 
 
 case, the product of any two terms, divided by the first term, is 
 equal to that term beyond the first, signified by the sum of their 
 indices. rp, JO, 1, 2, 3, 4, c., indices. 
 
 : \ 1, 3, 9, 27, 81, c., geometrical series. 
 Here, 4+3=7, the index of the 8th term. 
 81x27=2187, the 8th term, or 7th beyond the 1st, 
 
ALLIGATION MEDIAL. 187 
 
 4. Suppose a certain body, put in motion, should move 
 the length of one barleycorn the first second of time, one 
 inch the second, three inches the third second of time, 
 and so continue to increase its motion in triple proportion 
 geometrical ; how many yards would the said body move 
 in the space of half a minute 1 
 
 Ans. 953199685623yds. 1ft. lin. Ibar. j* which is no 
 less than five hundred and forty-one millions of miles. 
 
 ALLIGATION. 
 
 ALLIGATION teaches to mix several simples of different 
 qualities, so that the composition may be of a middle 
 quality ; and is commonly distinguished into two principal 
 cases, called Alligation Medial and Alligation Alternate. 
 
 ALLIGATION MEDIAL. 
 
 ALLIGATION MEDIAL is the method of finding the rate 
 of the compound, from having the rates and quantities f 
 the several simples given. 
 
 RULE. Multiply each quantity by its rate ; then divide 
 the sum of the products by the sum of the quantities, or 
 the whole composition, and the quotient will be the rate 
 of the compound required. 
 
 EXAMPLES. 
 
 L Suppose 20 bushels of wheat at 10s. per bushel, 36 
 bushels of rye at 6s. per bushel, and 40 bushels of barley 
 at 4s. per bushel, were mixed together; what would a 
 bushel of this mixture be worth ? 
 20x10=200 
 36 X 6=216 
 40x 4=160 
 
 96 )576(6s. Answer. 
 
 576 
 
 2. A composition being made of 5 pounds of tea at 7s. 
 per pound, 9 pounds at 8s. 6d. per pound, and 14^- pounds 
 at 6s. lO^d. per pound ; what is a pound of it worth ? 
 
 Ans. 7s. 4f-d.+ 
 
 3. A goldsmith mixes 8 pounds 5 ounces of gold of 
 14 carats fine, with 12 pounds 8 ounces of 18 ; what is 
 the fineness of this mixture ? Ans. IGy^y carat*. 
 
188 ALLIGATION ALTERNATE. 
 
 4. If with 40 bushels of corn at 4s. per bushel, there 
 are mixed 10 bushels at 6s. per bushel, 30 bushels at 5s. 
 per bushel, and 20 bushels at 3s. per bushel ; what will 
 10 bushels of that mixture be worth ? Ans. $7,16cts. 
 
 5. A grocer would mix 12cwt. of sugar at 10 dollars 
 percwt. with 3cwt. at 8f dollars per cwt. and 8cwt. at 
 7 dollars per cwt. ; what will a cwt. of this mixture be 
 worth ? Ans. gS,95cts. 6mill.+ 
 
 6. If 16 gallons of wine at 1 dollar 25 cents, and 4 
 gallons of water, be mixed with 40 gallons of wine at 3 
 dollars per gallon ; what will the mixture be worth per 
 gallon ? Ans. $2,33cts. 
 
 ALLIGATION ALTERNATE. 
 
 ALLIGATION ALTERNATE is the method of finding what 
 quantity of any number of simples whose rates are given, 
 will compose a mixture of a given rate; so that it is the 
 reverse of alligation medial, and may be proved by it. 
 
 PCULE. Write the rates of the simples in a column 
 under each other. 
 
 Connect, or link with a continued line, the rate of each 
 .simple, which is less than that of the compound, with one 
 or any number of those, that are greater than the com- 
 pound ; and each greater rate with one or any number 
 of the If, 
 
 Write the difference between the mixture rate, and that 
 of each of the fim pie, Opposite to the rates, with which 
 they are respectively linked. 
 
 Then, if only one difference stand against any rate, it 
 will be the quantity belonging to that rate; but if there 
 be several, their sum will be the quantity. 
 
 EXAMPLES. 
 
 1. A merchant would mix wines at 14s. 15s. 19s. and 
 
 22s. per irall. to that the mixture may be worth 18s. per 
 gallon ; what quantity of each must be taken ? 
 
 Or thus : 
 
 1 .... at 14s. 
 
 Ans. 
 
 J 3 .... at 22s. 
 
ALLIGATIOK ALTERNATE. 189 
 
 2. How much corn at 2s. 6d. 3s. 8d. 4s. and 4s. 8d. per 
 bushel, must be mixed together, that the compound may 
 be worth 3s. lOd. per bushel ? 
 
 Ans. 12 at 2s. 6d. 12 at 3s. 8d. 18 at 4s, &, 18 at 4s. 8d. 
 
 3. A goldsmith has gold of 18 carats fine, 16, 19, 22 
 and 24 ; how much must he take of each to make it 21 
 carats fine? . ( 3oz. of 16, loz. of 18, loz. of 19, 
 
 3 ' I 5oz. of 22, and 5oz. of 24 carats fine. 
 
 4. It is required to mix wine at 80 cents, wine at 70 
 cents, cider at lOcts. and water together, so that the mix- 
 ture may be worth 50cts. per gallon. 
 
 Ans. 9gals. of each sort of wine, 5 of cider &> 5 of water. 
 
 CASE 2. When the whole composition is limited to a cer- 
 tain quantity. 
 
 RULE. Find an answer as before by linking ; then 
 say, as the sum of the quantities, or differences thus de- 
 termined is to the given quantity, so is each ingredient 
 found, to the required quantity of each. 
 
 EXAMPLES. 
 
 1. How many gallons of water at Octs. per gallon, must 
 be mixed with wine worth 60cts. per gallon so as to fill 
 a cask of 100 gallons, and that a gallon may be afforded 
 at50cts.? 
 
 *n/ l 10 
 50 J 60 (50 
 
 60 : 100 : : 10 : 60 : 100 : : 50 : 
 10 50 
 
 I 
 
 6,0)100,0. 6,0)500,0 
 
 16f 83* 
 
 Ans. 16f gallons of water, and 83 of wine. 
 
 2. How much gold of 15, of 17, of 18 and 22 carats 
 fine, must be mixed together to form a composition of 40 
 ounces of 20 carats fine ? 
 
 Ans. 5oz. of 15, 17 and 18, and 25oz. of 22. 
 
 3. Wine at 3s. 6d. and at 5s. 9d. per gallon, is to be 
 mixed, so that a hogshead of 63 gallons may be sold for 
 <12 12s. ; how many gallons must be taken of each 1 
 
 Ans. 14gals. at 5s. 9d. and 49gals. at 3s. 6d. 
 
190 SINGLE POSITION. 
 
 CASE 3. When one of the ingredients is limited to a cer- 
 tain quantity. 
 
 RULE. Take the difference between each price and 
 the mean rate as before ; then, as the difference of that 
 simple whose quantity is given, is to the rest of the dif- 
 ferences severally, so is the quantity given to the several 
 quantities required. 
 
 EXAMPLES. 
 
 1. A grocer would mix teas at 1 dollar 20cts. 66cts. 
 and 1 dollar per pound, with 20 pounds at 40 cents per 
 pound ; how much of each sort must he take to make the 
 composition worth 80 cents per pound ? 
 
 ife tfc fb tb 
 
 40 : 20 : : 20 : 10 at 66cts. ) 
 40: 14:: 20: 7 at $1 VAns. 
 40:40: : 20: 20 at 1,20 J 
 
 low much wine at SOcts. at 88 and 92 per gallon, 
 must be mixed with 4 gallons at 75cts. per gallon, so 
 that the mixture may be worth 86 cents per gallon ? 
 
 Ans. 4gals. at SOcts. 8^ at 88 and 8J- at 92. 
 3. With 95 gallons of wine at 8s. per gallon, I mixed 
 other wine at 6s. 8d. per gallon, and some water ; then I 
 found it stood me in 6s. 4d. per gallon ; I demand how 
 much wine at 6s. 8d. I took, and how much water. 
 Ans. 95 gallons wine at 6s. 8d. and 30 gallons water. 
 
 POSITION. 
 
 POSITION is a method of performing such questions as 
 cannot be resolved by the common direct rules, and is of 
 two kinds, Single and Double. 
 
 SINGLE POSITION. 
 
 Single Position teaches to resolve those questions 
 whose results are proportioned to their suppositions. 
 
 RULE. 1. Take any number and perform the same 
 operations with it, as are described to be performed in 
 the question. 
 
DOUBLE POSITION. 101 
 
 2. Then say, as the result of the operation is to the 
 position, so is the result in the question to the number re- 
 quired. 
 
 EXAMPLES. 
 
 1. A's age is double that of B, and B's is triple that of 
 C, and the sum of all their ages is 140 ; what is the age 
 of each ? Suppose A's age to be 48 
 Then will B's =4=24 
 And C's=- 2 ^=8 
 
 80 sum. 
 
 As 80 : 48 :: 140 : 84 = A's age. 
 Consequently % 4 -=42=B's. 
 
 140 Proof. 
 
 2. A certain sum of money is to be divided between 4 
 persons in such a manner that the first shall have ^ of it, 
 the second |, the third , and the fourth the remainder, 
 which is 28 dollars ; what is the sum ? Ans. $112. 
 
 3. A person, after spending J- and of his money, had 
 60 dollars left ; what had he at first ? Ans. $144. 
 
 4. What number is that which being increased by , ^, 
 and of itself, the sum will be 125 7 Ans. 60. 
 
 5. A person lent his friend a sum of money, to receive 
 interest for the same at 6 per cent, per annum, simple 
 interest ; at the end of three years he received for princi- 
 pal and interest 383 dollars 50 cents ; what was the sum 
 lent ? Ans. 325 dollars. 
 
 6. A cistern is supplied with three cocks, A, B, and C : 
 A can fill it in 1 hour, B in 2, and C in 3 ; in what time 
 will it be filled by all of them together 1 Ans. T 6 T hour. 
 
 DOUBLE POSITION. 
 
 DOUBLE POSITION teaches to resolve questions by mak- 
 ing two suppositions of false numbers.* 
 
 * Questions in which the results are not proportional to their 
 positions, belong 1 to this rule ; such are those, in which the num- 
 ber sought is increased or diminished by some given number, 
 which is no known part of the number required. 
 
192 DOUBLE POSITION. 
 
 RULE. 1. Take any two convenient numbers, and 
 proceed with each according to the conditions of the 
 question. 
 
 2. Find how much the results are different from the 
 result in the question. 
 
 3. Multiply each of the errours by the contrary suppo- 
 sition. 
 
 4. If the errours be alike, divide the difference of the 
 products by the difference of the errours, and the quo- 
 tient will be the answer. 
 
 5. If the errours be unlike, divide the sum of the pro- 
 ducts by the sum of the errours, and the quotient will be 
 the answer. 
 
 NOTE. The errours are said to be alike, when they are 
 both too great, or both too little ; and unlike, when one is 
 too great, and the other too little. 
 
 EXAMPLES. 
 
 1. A lady bought cambric for 40 cents a yard, and In- 
 dia cotton at 20 .cents a yard ; the whole number of yards 
 she bought was 8, and the whole cost 2 dollars ; how 
 many yards had she of each sort ? 
 
 Suppose 4 yards of cambric, value $l,60cts. 
 Then she must have 4 yards of cotton, value 80 
 
 Sum of their values, 2,40 
 So that the first errour is-f 40 
 
 Again, suppose she had 3 yards of cambric, $l,20cts. 
 Then she must have 5 yards of India cotton 1,00 
 
 Sum of their values, 2,20 
 So that the second errour is +20 
 Then 40 20 =20= difference of the errours. 
 Also 4x20=80=product of the first supposition and 
 
 second errour. 
 And 3x40=120=product of the second supposition 
 
 and first errour. 
 
 And 120 80=40=their difference. 
 Whence 40^-20=2 yards of cambric. > ^^ 
 And 8 2=6 yards of India cotton. / 
 
 2. A and B have both the same income ; A saves of 
 his yearly ; but B, spending 50 dollars a year more than 
 
PERMUTATION AND COMBINATION. 193 
 
 A, at the end of 4 years is 100 dollars in debt ; what is 
 their income, and what do they spend per annum ? 
 
 ( Their income is $125 per year. 
 Ans.J A spends $100. 
 
 (B spends 8150. 
 
 3. A labourer was hired for 40 days upon these condi- 
 tions, that he should receive 2 dollars for every day he 
 wrought, and forfeit 1 dollar for every day he was idle ; 
 at the expiration of the time he was entitled to 50 dollars ; 
 how many days did he work, and how many was he idle ; 
 
 Ans. He wrought 30 days, and was idle 10. 
 
 4. A man had 2 silver cups of unequal weight, with 1 
 cover for both, weight 5oz ; now if he put the cover on 
 the less cup, it will be double the weight of the greater ; 
 and put on the greater cup, it will be three times the 
 weight of the less cup ; what is the weight of each cup? 
 
 Ans. 3oz. the less, and 4oz. the greater. 
 
 5. A person being asked what o'clock it was, an- 
 swered that the time past from noon was equal to -% of 
 the time to midnight ; required the time. 
 
 Ans. 36 minutes past 1. 
 
 6. There is a fish whose head is ten feet long ; his tail 
 is as long as his head and half the length of his body, 
 and his body is as long as his head and tail ; what is the 
 whole length of the fish ? Ans. 80 feet. 
 
 7. A and B laid out equal sums of money in trade ; A 
 gained a sum equal to of his stock, and B lost 225 dol- 
 lars ; then A's money was double that of B's ; what did 
 each lay out. Ans. 
 
 PERMUTATION AND COMBINATION. 
 
 THE permutation of quantities is the showing how 
 many different ways the order or position of any given 
 number of things may be changed. 
 
 The combination of quantities is the showing how of- 
 ten a less number of things may be taken out of a great- 
 er, and combined together, without considering their pla- 
 ces, or the order in which they stand. 
 R 
 
Proof. 
 
 194 PERMUTATION AND COMBINATION 
 
 PROBLEM 1. 
 
 To find the number of permutations, or changes, that 
 can be made of any number of things, all differing from 
 each other. 
 
 RULE. Multiply all the terms of the natural series of 
 numbers, from one up to the given number, continually 
 together, and the last product will be the answer. 
 
 EXAMPLES. 
 
 1. How many changes may be made with these three 
 letters, A, B, C. 
 
 CHANGES. 
 
 1 a b c 
 
 2 a c b 
 b a c 
 
 2 be a 
 
 3 cab 
 
 c b a 6 J 
 6 Answer. 
 
 2. How many changes may be rung upon 6 bells ? 
 
 Ans. 720. 
 
 3. How many changes may be rung upon 12 bells, and 
 how long would they be ringing but once over, supposing 
 10 changes might be rung in one minute, and that the 
 year contains 365 days, 6 hours 1 
 
 . i 479001600 changes, and 91 
 5 ' \ years 3w. 5d. and 6 hours. 
 
 4. A young scholar coming into a town for the conven- 
 ience of a good library, demanded of the gentleman with 
 whom he lodged, what his diet would cost for a year ; he 
 told him $150; but the scholar not being certain what 
 time he should stay, asked him what he should give him 
 for so long as he could place his family (consisting of 6 
 persons beside himself) in different positions every day at 
 dinner ; the gentleman told him 850 ; to this the scholar 
 agreed what time did he stay 1 Ans. 41)40 days. 
 
 PROBLEM II. 
 
 Any number of different things being given, to find 
 how many changes can be made out of them, by taking 
 any given number at a time. 
 
PERMUTATION AND COMBINATION. 195 
 
 RULE. Take a series of numbers, beginning at the 
 number of things given, and decreasing by one, till the 
 number of terms be equal to the number of things to be 
 taken at a time, multiply these terms into each other ; 
 and the product will be the answer. 
 
 EXAMPLES. 
 
 1. How many changes may be made out of the three 
 letters a, b, c by taking two at a time 1 
 CHANGES. 
 
 3 a b 
 
 2 ba 
 
 a c 
 
 6 Answer. c a 
 
 be 
 
 cb 
 
 ^ Proof. 
 
 2. How many changes may be made with the nine di- 
 gits, by taking 3 at a time ? Ans. 504. 
 
 3. How many words may be made with the alphabet by 
 taking five letters at a time, supposing that a number of 
 consonants may make a word ? Ans. 5100480. 
 
 PROBLEM III. 
 
 To find the compositions of any number, in an equal 
 number of sets, the things themselves being all different. 
 
 RULE. Multiply the number of things in every set 
 continually together, and the product will be the answer, 
 
 EXAMPLES. 
 
 1. Suppose there are four companies, in each of which 
 there are nine men; it is required to find how many ways 
 four men may be chosen, one out of each company. 
 
 9x9x9x9=6561 the answer. 
 
 ^. How many changes are there in throwing five dice ? 
 
 Ans. 7776. 
 
 3. Suppose there are four companies, in one of which 
 there are 6 men, in another 8, and in each of the other 
 two 9 ; what are the choices by a composition of four 
 men, one out of each company 1 Ans. 3888. 
 
 4. Suppose a man undertakes to throw an ace, at one 
 throw, with 4 dice ; what is the probability of his effect* 
 ing it ? Ans. as 671 to 625, 
 
196 MISCELLANEOUS QUESTIONS. 
 
 MISCELLANEOUS QUESTIONS. 
 
 1. A gentleman bought 27 yards of cloth at 2s. per 
 yard, 24 yards at 3s. l^d. per yard, 25 yards at Is. 8d. 
 per yard ; he also bought 3 yards of broadcloth, the price 
 of which he does not recollect ; but on counting his 
 money he found he had expended ^11 19s. 2d. ; what 
 did his broadcloth cost per yard, in Federal Money ? 
 
 Ans. $3,75cts. 
 
 2. A servant went to market with <5, and bought eggs 
 at 7 for 4d. ; 2 pair of fowls at 2s. 4d. a pair ; 17 pigeons 
 at 3s. per dozen ; 3 rabbits at I4d. each; and 3 dozen of 
 larks at 14d. per dozen ; he also paid the baker ,2 17s. 
 Id. ; when he returned he had 21s. left ; how many eggs 
 did he buy ? Ans. 126. 
 
 3. I have a drawer 17 inches long, 12 inches broad and 
 7 inches deep ; how many one inch dice will it hold 1 
 
 Ans. 1428. 
 
 4. At a certain election 375 persons voted, and the 
 candidate chosen had a majority of 91 : how many voted 
 for each ? Ans. 233 and 142. 
 
 5. Suppose a man to step 30 inches at a time, and to 
 go 4 miles an hour ; how many times does he step in a 
 minute ? Ans. 140f . 
 
 6. The divisor is 43967, the quotient 2737226, and the 
 remainder 27672 ; what is the dividend ? 
 
 Ans. 120347643214. 
 
 7. A prize of $1000 is to be divided between two per- 
 sons whose shares are in proportion of 7 to 9 ; required 
 the share of each. A ( S437,50cts. 
 
 s * \ g562,50cts. 
 
 8. After paying away and -^ of my money, I had 66 
 guineas left in my purse ; what was in it at first ? 
 
 Ans. 120. 
 
 9. A reservoir for water has two cocks to supply it ; 
 the first alone will fill it in 40 minutes, the second in 50 
 minutes and it has a discharging cock by which it may 
 be emptied, when full, in 25 minutes. Now supposing 
 that these three cocks are all opened, that the water comes 
 in, and that the influx and the efflux of the water are al- 
 ways alike., in what time would the cistern be filled ? 
 
 Ans. 3 hours, 20 minutes* 
 
MISCELLANEOUS QUESTIONS. 197 
 
 10. In the latitude of Hallowell, a degree of longitude 
 measures about 49 miles, 6 furlongs, and 1 1^ poles ; now 
 as the earth turns round in about 23 hours, 56 minutes, 
 at what rate per hour is the town of Hallowell carried by 
 this motion from west to east 1* Ans. 748-i^J miles. 
 
 11. If the earth turns round in 23 hours 56 minutes, at 
 what rate per hour are the inhabitants of the city of Quito 
 in South America, which lies under the equator, carried 
 from west to east by this rotation 1 Ans. 1045^1 miles. 
 
 12. In a mixture of wine cider, ^ of the whole added 
 to 25 gallons, was wine ; and 4- part, less 5 gallons, was 
 cider ; how many gallons were there of each ? 
 
 Ans. 85 of wine and 35 of eider. 
 
 1&. A hare is 50 of her own leaps before a greyhound, 
 and takes 4 leaps to the greyhound's 3 ; but two of the 
 greyhound's leaps are as much as 3 of the hare's : how 
 many leaps must the greyhound take to catch tbe hare 1 
 
 Ans. 300. 
 
 14. Out of a cask of wine which trad leaked away J- 
 part, 21 gallons were drawn ; and then 'being gauged, it 
 was found to be half full ; how many gallons did it hold t 
 
 Ans, 126. 
 
 15. What part of 4d. is f of 6 pence ? Ans. f. 
 
 16. WJiat number is that from which, if 5 be subtract- 
 ed f of the remainder is 80 ? Ans. 125. 
 
 17. A post is'J- in the mud, -J in the water, and 10 feet 
 above the water : what is its whole length ? Ans. 24. 
 
 18. A captain, mate, and 20 seamen, took a prize worth 
 $3501 ; of which the captain takes 11 shares, and the 
 mate 5 shares ; the remainder of the prize is equally di- 
 vided among the sailors ; how much did each man re- 
 ceive ? A | The Capt. $1069,75cts. The mate 
 
 5,25cts. Each sailor $97,25 cents. 
 
 19. A stationer sold quills at $1, 83cts. per thousand, 
 by which he cleared | of the money ; "but as they grew 
 scarcer, he raised them to #2,25cts. per thousand ; 
 what did he clear per cent, by the latter price ? 
 
 Ans. $96,36 T 4 T cts. gained per cent, by the latter price, 
 
 20. Bought a quantity of goods for $250, aud 3 month* 
 
 *The earth moves 1 degree in 3* 59* 2G**f of 
 real time, or in 4* of solar or tropical time.. 
 B2 
 
198 MISCELLANEOUS QUESTIONS. 
 
 after sold them for $275 ; how much per cent, per annum 
 did I gain by them ? Ans. $40. 
 
 21. In what time will the interest of g72,60cts equal 
 that of $15,25cts. for 04 days, at any rate of interest ? 
 
 Ans. 13f days. 
 
 22. What sum of money will amount to $132,81cts. 
 2jm. in 15 months, at 5 per cent, per annum simple in- 
 terest ? Ans. $125. 
 
 23. A person possessed of of a ship sold f of his 
 share for $1260; what was the reputed value of the 
 whole at the same rate ? Ans. 5040. 
 
 24. Of my T % of a farm I sell f of f ; what I then own 
 is worth 8185 ; what is that part, and what the value of 
 the farm ? Ans. ^ and lhe farm $1200. 
 
 25. What number is that to"which if T 3 ^ of 4^-of f be 
 added, the total will be one ? Ans. f f f f . 
 
 26. If of | of I of a ship be worth of f of | of the 
 cargo, valued at 40000 dollars ; what is the value of the 
 ship and cargo ? Ans. $50744,81 cts.+ 
 
 27. A grocer would mix a quantity of sugar at lOd. per 
 pound, with other sugars at 7d. 5d. and 4d. per pound, 
 intending to make up a compound worth 6d. per pound ; 
 what quantity of each must he take? Ans l^ffo at lOd. 
 
 Itb at 7d. ltb at 5d. and 4ft at 4d. 
 
 28. If 1000 men besieged in a town,with provisions for 
 5 weeks, allowing each man 16 ounces a day, were rein- 
 forced with 500 men more, and hear that they cannot be 
 relieved till the end of 8 weeks ; kow many ounces a day 
 must each man have, that the provisions may last that 
 time ? Ans. 6| ounces. 
 
 29. Sound, not interrupted, is found by experiment to 
 move uniformly about 1150 feet in a second of time ; how 
 long then,after firing an alarm gun at Fort Independence* 
 may the same be heard at Cambridge, taking the distance 
 at 5f miles ? Ans. 26f seconds. 
 
 30. If I see the flash of a gun fired by a vessel in dis- 
 tress at sea, which happens, we will suppose, at the instant 
 of its going off, and hear the report a minute and 3 sec- 
 onds afterwards ; how far is she off! Ans. 72450 feet. 
 
 SI* An elm plank is 14 feet, 3 inches long ; what dis- 
 tance from the edge must a line be struck to take off a 
 yard square 1 Ans, 7| inches. 
 

 MISCELLANEOUS QUESTIONS. 199 
 
 32. A man dying left his wife in expectation that a 
 child would be afterwards added to the family, and in 
 making his will ordered, that if the child were a son, f of 
 his estate should belong to him, and the remainder to his 
 mother ; but if it were a daughter, he appointed the 
 mother f and the child the remainder ; but it happened 
 that the addition was both a son and a daughter, by which 
 the widow lost in equity 2400 dollars more than if there 
 had been only a girl ; what would have been her dowry, 
 had she had only a son ? Ans. $2100. 
 
 33. Having a piece of land 11 perches in breadth, I 
 demand what length of it must be taken to contain an 
 acre, when four perches in breadth require 40 perches in 
 length to contain the same ? Ans. 14per. 3yds. 
 
 34. If a gentleman whose annual income is ^1000 9 
 spends 20 guineas, each 21s., a week, will he fall in debt, 
 or save money, and how much in the year ? 
 
 Ans. ,92 in debt. 
 
 35. What sum of money will produce as much interest 
 in 3 years, as $210,15cts. can produce in 5 years and 5 
 months? Ans. $350,25cts. 
 
 36. If $100 in 5 years be allowed to gain 820, 50cts. 
 in what time will any sum of money double itself, at the 
 fame rate of interest ? Ans. 24^f years. 
 
 37. What difference is there between the interest of 
 $350 at 4 per cent, for 8 years, and the discount of the 
 same sum, at the same rate, and for the same time ? 
 
 Ans. $27,15^cts. 
 
 38. If by selling goods at $2 per cwt. I gain 20 per 
 cent., what do I gain or lose per cent, by selling at $2 
 per cwt. 1 Ans. $8 per cent. gain. 
 
 39. Required the length of a shore, which, strutting 
 11 feet from the upright of a building, may support a 
 jamb 23 feet, 9 inches from the ground. Ans. 26ft 2i.+ 
 
 40. A clears $13 in 6 months, B $18 in 5, and C g23 
 in 9, his stock being g72 ; what, then, is the general 
 stock? Ans. $236,09f. 
 
 41. A person making his wiff, gave to one child $ of 
 his estate, and the rest to another ; when these legacies 
 came to be paid, the one turned out $600 more than the 
 other : what did the testator die worth? Ans. $2000. 
 
 42. A father devised -f^ of his estate to one of his sons 
 and ^ of the residue to another, and the remainder to bis 
 
200 MISCELLANEOUS QUESTIONS. 
 
 widow for life ; the children's legacies were found to be 
 S257,16f cts. different : pray what sum did he leave the 
 widow the use of? Ans. $635,04 T 6 4 2 r cts. 
 
 43. A had 12 pipes of wine, which he parted with to 
 B at 4^ per cent, profit, who sold them to C for S40, 
 GOcts. advantage ; C made them over to D for $605,50cts., 
 and cleared thereby 6 per cent. : how much a gallon did 
 this wine cost A ? Ans. SS^^fyetB. 
 
 44. Laid out $I65,75cts. in wine at gleets, a gallon ; 
 some of which receiving damage in carriage, I sold the 
 rest at Slfcts. a gallon, which produced only $110,83^ 
 cts. ; what quantity was damaged 1 Ans. 430gals 
 
 45. A young hare starts 40 yards before a greyhound, 
 and is not perceived by him till she has been up 40 sec- 
 onds ; she scuds away at the rate of ten miles an hour, 
 and the dog, on view, makes after her at the rate of 18 ; 
 how long will the course hold, and what ground will be 
 run over by the dog ? Ans. 60^-sec. and 530 yds. run. 
 
 46. If I leave Hallowell at 8 o 'clock on Monday morn- 
 ing for Newburyport, and ride at the rate of 3 miles an 
 hour without intermission; and B sets out from Newbu- 
 ryport for Hallowell at 4 o'clock the same evening, and 
 rides 4 miles an hour constantly: supposing the distance 
 between the two towns to be 130 miles, whereabout on 
 the road shall we meet ? 
 
 Ans. 69f miles from Hallowell, which will be in Saco. 
 
 47. X, Y, and Z, can, working together, complete a 
 staircase in 12 days ; Z is man enough to do it alone in 
 24 days, and X, in 34; in what time, then, could Y get 
 it done himself? Ans. 8 If days. 
 
 48. A and B together can build a boat in 18 days, and 
 with the assistance of C, they can do it in 11 days; in 
 what time then, would C do it by himself? 
 
 Ans. 28f days. 
 
 49. Laid out in a lot of muslin ^500^ upon examina~ 
 tion of which 3 parts in 9 proved damaged, so that I 
 could make but 5s. a yard of the same ; and by so doing, 
 find 1 lost =50 by it ; at what rate per ell Eng. am I to 
 part with the undamaged muslin, in order to gain 50 
 pon the whole ? Ans. 11s. 7fd. 
 
 60. If the sun move, every day, one degree, and the 
 sxioon thirteen ; and, at a certain time, the sun be ai the 
 
MEASUREMENT OF GRINDSTONES. 201 
 
 beginning of Cancer, and, in three days after, the moon 
 at the beginning of Aries ; the place of their next follow- 
 ing conjunction is required. Ans. 10 45' of Cancer. 
 
 51. A person being asked the time of day, answered, 
 it is between 4 and 5 ; but a more particular answer be- 
 ing required, he said, that the hour and minute hands 
 were then exactly together ; what was the time 7 
 
 Ans. 21 T 9 T min. past 4. 
 
 52. What weight, hung at 70 inches' distance from the 
 fulcrum of a steelyard, will equiponderate a hhd. of to- 
 bacco, weighing 950&. freely suspended at 2 inches' dis- 
 tance on the contrary side! Ans. 27ft. 2oz. 4^drs. 
 
 53. If two places lie so much due east and west of each 
 other, that it is found, by observation, to be noon at the 
 former 2 hours, 6 min. and 30 seconds sooner than at the 
 latter ; how many degrees are they apart 1 
 
 Ans. 31 37' 30 seconds. 
 
 54. If Paris, in France, be in 2 20' east longitude 
 from Greenwich, and Hallowell in 69 42' west longitude 
 from Greenwich ; when it is noon at Paris, what time of 
 day is it at Hallowell ? 
 
 Ans. 7h. 1 1m. 52s. in the morning. 
 
 MEASUREMENT OF GRINDSTONES. 
 
 GRINDSTONES are sold by the stone, and their contents 
 found as follows ;* 
 
 RULE. To the whole diameter add half of the diam- 
 eter, and multiply the sum of these by the same half, and 
 this product by the thickness ; divide this last number by 
 1728, and the quotient is the contents, or answer re- 
 quired. 
 
 EXAMPLES. 
 
 1. What are the contents of a grindstone 24 inches 
 diameter, and 4 inches thick 7 
 
 24+12x12x4 
 
 =1 stone. Ans. 
 
 1728 
 
 2. What are the contents of a grindstone 36 inches 
 diameter, and 4 inches thick : Ans. 2 stone. 
 
 * 24 inches in diameter, and 4 inches thick, make a stone. 
 
202 MENSURATION. 
 
 MENSURA TION of Superficies and Solids. 
 Section 1. OF SUPERFICIES. 
 
 Superficial measure is that, which relates to length and 
 breadth only, not regarding thickness. It is made up of 
 squares, either greater or less, according to the different 
 measures by which the dimensions of the figure are taken 
 or measured. Land is measured by this measure, its di- 
 mensions being usually taken in acres, rods, and links. 
 The contents of boards, also, are found by this measure, 
 their dimensions being taken in feet and inches. Be- 
 cause 12 inches in length make 1 foot of long measure, 
 therefore 12x12 = 144, the square inches in a superficial 
 foot, &c. 
 
 CASE 1. To find the Area of a square having equal sides. 
 RULE. Multiply the side of the square into itself, and 
 the product will be the area, or superficial content, of 
 the same name with the denomination taken, whether 
 inches, feet, yards, rods and links, or acres. 
 
 EXAMPLES. 
 
 1. How many square feet of boards are contained in 
 the floor of a room which is 20 feet square ? 
 
 20x20=400 feet, the Answer. 
 
 2. Suppose a square lot of land measures 26 rods on 
 each side, how many acres does it contain 1 
 
 26x26 
 
 As 160 square rods make an acre : therefore zi=4ac. 36 ro. 
 
 160 Ans. 
 
 CASE 2. To measure a parallelogram or long square. 
 
 RULE. Multiply the length by the breadth, and the 
 product will be the area, or superficial content, in the 
 same name as that, in which the dimension was taken, 
 whether inches, feet, or rods, &c. 
 
 EXAMPLES. 
 
 1. A certain garden, in form of a long square, is 96 feet 
 long, and 54 feet wide ; how many square feet of ground 
 are contained in it? 96x54=5184 square feet. Ans. 
 
 2. A lot of land, in form of a long square, is 120 rods 
 in length, and 60 rods wide ; how many acres are in it ? 
 120x60=7200 sq. rods. And 7200-i-160=45 acres. Ans. 
 
OF SUPERFICIES. 203 
 
 3. How many acres are in a field of oblong form, 
 whose length is 14,5 chains, and breadth 9,75 chains ? 
 
 Ans. 14ac. Oroo. 22rods. 
 
 NOTE. The Gunter's chain is 66 feet, or 4 rods, long, 
 and contains 100 links. Therefore, if dimensions be giv- 
 en in chains and decimals, point off from the product one 
 more decimal place than are contained in both factors 
 and it will be acres and decimals of an acre ; if in chains 
 and links, do the same, because links are hundredths of 
 chains, and, therefore, the same as decimals of them. 
 Or, as 1 chain wide, and 10 chains long, or 10 square 
 chains, or 100000 square links, make an acre, it is the 
 same as if you divide the links in the area by 100000. 
 
 4. If a board or plank be 21 feet long, and 18 inches 
 broad, how many square feet are contained in it ? 
 
 18 inches =1,5 foot. And 21 x 1,5=3 1,5 feet. Ans. 
 
 Or, in measuring boards, you may multiply the length 
 in feet by the breadth in inches, and divide the product 
 by 12 ; the quotient will give the answer in square feet, 
 &c. 21x18 
 
 Thus, in the preceding example, =31^- sq. feet 
 
 as before. 12 
 
 5. If a board be 8 inches wide, how much in length 
 will make a foot square ? 
 
 RULE. Divide 144 by the width ; thus, 8)144 
 
 Ans. 18 inch. 
 
 6. If a piece of land be 5 rods wide, how many rods 
 in length will make an acre 1 
 
 RULE. Divide 160 by the width, and the quotient will 
 be the length required ; thus, 
 5)160 
 
 32 rods in length. Ans. 
 
 NOTE. When a board, or any other surface, is wider 
 at one end than the other, but yet is of a true taper, 
 you may take the breadth in the middle, or add the 
 width of both ends together, and halve the sum, for the 
 mean width : then multiply the said mean breadth in 
 either case, by the length ; the product is the answer, or 
 area sought. 
 
204 MENSURATION 
 
 7. How many square feet in a board 10 feet long, and 
 13 inches wide at one end, and 9 inches wide at the 
 other? 13+9 
 
 =!! inches mean width. 
 
 2 ft. in. 
 
 10x11 
 
 =9|ft. Ans. 
 
 12 
 
 8. How many acres are in a lot of land which is 40 
 rods long, and 30 rods wide at one end, and 20 rods wide 
 at the other 1 30+20 
 
 =25 rods mean width. 
 
 2 Then, 25x40 
 
 =61 acres. Ans. 
 
 160 
 
 9. If a farm lie 250 rods on the road, and, at one end, 
 be 75 rods wide, and, at the other, 55 rods wide, how 
 many acres does it contain ? Ans. 10 lac. 2roo. lOro. 
 
 CASE 3. To measure the surface of a triangle. 
 
 Definition. A triangle is any three cornered figure 
 which is bounded by three right lines.* 
 
 RULE. Multiply the base of the given triangle into 
 half its perpendicular height, or half the base into the 
 whole perpendicular, and the product will be the area. 
 
 EXAMPLES. 
 
 1. Required the area of a triangle whose base or longest 
 side is 32 inches, and the perpendicular height 14 inches. 
 14~t~2=37= the perpend, and 32x7=224 sq. in. Ans. 
 
 2. There is a triangular or three cornered lot of land, 
 whose base or longest side is 51 rods ; the perpendicu- 
 lar, from the corner opposite to the base, measures 44 
 rods ; how many acres does it contain ? 
 
 44-r-2=22= ha If perpendicular. 
 And 51,5x22 
 
 =7 acres, 13 rods. Ans. 
 
 160 
 
 * A triangle may be either right-angled or oblique ; in either 
 case, the teacher can easily give the scholar a just idea of the 
 base and perpendicular, by marking it down on a slate or paper, 
 &c. In a right-angled triangle, the longest of the two legs which 
 include the right-angle, is called the base ; but in such as are 
 oblique, the longest of the three sides is so called. 
 
ON SUPERFICIES. 
 
 205 
 
 3. If a piece of land lie in the form of a right-angled 
 triangle, its base being 37 rods, and the perpendicular 
 line being 24 rodfc, how many acres are in it 1 
 
 Ans. 2,8617+ acres. 
 
 4. If the base of a triangular field be 7 chains and 50 
 links, and the perpendicular 4 chains and 25 links, how 
 much does it contain ? Ans. lac. 2roo. 15rods. 
 
 Joists and Planks are measured by the following 
 RULE. Find the area of one side of the joist or plank, 
 by one of the preceding rules ; then multiply it by the 
 thickness in inches ; and the last product will be the su- 
 perficial content. 
 
 EXAMPLES. 
 
 1. What is the area, or superficial content, or board 
 measure, of a joist, 20 feet long, 4 inches wide, and 3 in- 
 ches thick? 20x4 
 
 X3=20 feet. Ans. 
 
 12 
 
 2. If a plank be 32 feet long, 17in. wide, and Sin. thick, 
 what is the board measure of it ? Ans. 136 feet. 
 
 NOTE. There are some numbers, the sum of whose squares 
 makes a perfect square ; such are 3 and 4, the sum of whose 
 squares is 25, the square root of which is 5 ; consequently, when 
 one leg- of a right-angled triangle is 3, and the other 4, the hy- 
 potenuse must be 5. And if 3, 4, and 5, be multiplied by any 
 other numbers each by the same, the products will be sides of 
 true right-angled triangles. Multiplying them by 2, gives 6, 8, 
 and 10 by 3, gives 9, 12 and 15 by 4, gives 12, 16 and 20, &c. ; 
 all which are sides of right-angled triangles. Hence architects 
 in setting off the corners of buildings, commonly measure 6 feet 
 on one side, and 8 feet on the other ; then, laying 1 a 10 feet pole 
 across from those two points, it makes the corner a true right- 
 angle. 
 
 RULE 2. To Jind the, area of any triangle when the three 
 
 sides only are given. 
 
 RULE. From half the sum of the three sides subtract 
 each side severally ; multiply these three remainders and 
 the said half sum continually together ; then the square 
 root of the last product will be the area of the triangle. 
 
 EXAMPLE. 
 
 Suppose I have a triangular fish-pond, whose three sides 
 measure 400, 348, and 312yds. ; what quantity of ground 
 does it cover 1 Ans. 10 acres, 3roods, 8-f-rods. 
 
 S 
 
206 MENSURATION 
 
 CASE 4. To measure irregular surfaces. 
 RULE. Divide the figure or plane into triangles, by 
 drawing diagonal lines from one angle to another ; then 
 measure all the triangles, by either of the rules in Case 
 3 ; and the sum of their several areas will be the area of 
 the given figure. 
 
 EXAMPLE. 
 
 If a piece of ground be divided into two triangles by 
 a diagonal line drawn through it measuring 30 rods, and 
 two perpendiculars be let fall, one measuring 8 rods, and 
 the other 14 rods ; how many acres does it contain ? 
 
 Ans. 2* acres. 
 
 CASE 5. To measure a circle. 
 
 Definition. A circle is a figure bounded by a curve or 
 circular line, every part of which is equally distant from 
 the middle or centre. The curve line is called the per- 
 iphery or circumference ; a line drawn, from one side to 
 the other, through the centre, is called the diameter ; and 
 a line drawn, from the centre to the circumference, is call- 
 ed the semidiameter, (half diameter,) or radius. 
 
 PROBLEM I. 
 
 The diameter given to Jind the circumference. 
 RULE. As 7 are to 22, so is the given diameter to the 
 circumference ; or, more exactly, as 113 are to 355, so is 
 the diameter to the circumference, &c. 
 
 EXAMPLES. 
 
 1. What is the circumference of a wheel whose diam- 
 eter is 4 feet ? 
 
 As 7 : 22 : : 4 : 12,57+feet the circumference. Ans. 
 
 2. What is the circumference of a circle whose diam- 
 eter is 35 rods ? As 7 : 22 : : 35 : 110 rods. Ans. 
 
 NOTE. To find the diameter, when the circumference 
 is given, reverse the foregoing rule, and say, as 22 are to 7, 
 so is the given circumference to the required diameter ; or, 
 as 855 are to 11 3, so is the circumference to the diameter. 
 
 3. What is the diameter of a circle whose circumfer- 
 ence is 110 rodsl 
 
 As 23 : 7 : : 110 : 35 rods the diam. An. 
 
OF SUPERFICIES. 207 
 
 PROBLEM. II. 
 To find the area of a circle. 
 
 RULE. Multiply half the diameter by half the circum- 
 ference, and the product is the area ; or, if the diameter 
 alone is given, multiply the square of the diameter by 
 ,785398 or, which is near enough, by ,7854 and the 
 product will be the area. 
 
 EXAMPLES. 
 
 1. Required the area, or superficial content of a circle 
 whose diameter>is 12 rods, and circumference 37,7 rods, 
 18,85=half the circumference. 
 6=half the diameter. 
 
 113,10 area in square rods. Ans. 
 
 2. What is the superficial content of a circular garden 
 whose diameter is 11 rods I 
 
 By the second method. 
 
 11x11=121. ,7854x121=95,0334 rods. Ans. 
 
 3. What will be the cost of a circular platform to the 
 curb of a round well, at 10^ cents per square foot ; if the 
 diameter of the well be 42 inches, and the breadth of the 
 platform be 14 inches 1 Ans. $l,S7^cts.+ 
 
 PROBLEM III. 
 
 To find the area of a circle when the circumference alone is given. 
 
 RULE. Multiply the square of the circumference by 
 ,079577, or, which is near enough, by ,07958 and the 
 product will he the area. 
 
 NOTE. ,785398 is the area of a circle whose diameter 
 is 1, and ,079577 is the area of a circle whose circumfer- 
 ence is 1. 
 
 EXAMPLE. 
 
 What is the area of a circle whose circumference is 30 
 rods? 30 x30x, 07958 =71, 62200 rods. Ans. 
 
 PROBLEM IV. 
 
 The area of a circle given to Jind the diameter. 
 RULE. Divide the area by ,7854 and the square root 
 of the quotient is the required diameter* 
 
208 
 
 MENSURATION 
 
 EXAMPLES. 
 
 1. Required the diameter of a circle that, will contain 
 within its circumference the quantity of an acre of land. 
 1 acre=^4840sq, yds. Then 1/,4-ff J=88,5+yda. Ans - 
 
 2. In the midst of a meadow abounding with feed, 
 For two acres, to tether my horse, I've agreed ; 
 How long must the rope be, that, feeding all round, 
 He may n't graze less or more than the two acres of 
 
 ground ? Ans. 55^+yards. 
 
 3. A, B, and C, join to buy a grindstone, 36 inches in 
 diameter, which cost $3, and towards which A paid $1, 
 B, $1^, and C, 83^cts. The waste bole for the spindle 
 was 5 inches square. To what diameter ought the stone 
 to be worn, when B and C begin severally to work with 
 it allowing for the hole, and A first grinding down his 
 share, next B, and then C ? 
 
 j^ ( 29,324+inch. diameter where B begins to 
 
 ~\ grind; and 19,013+in. diam. C begins. 
 NOTE. Twice the square of the side of a square, will 
 bet he square of the diameter of its circumscribing circle. 
 
 PROBLEM V. 
 
 The area of a circle given to jftnd the circumference. 
 RULE. Divide the given area by ,07958 and the 
 square root of the quotient is the required circumference. 
 
 EXAMPLES. 
 
 1. The expense of turfing a round plot, at 4 pence per 
 square yard, was 2. 9s. 9|d. ,8 ; what was its circum- 
 ference 1 Ans. 130+feet. 
 
 2. How many feet of boards will fence a round garden, 
 containing just two acres, the fence five feet high; and 
 what will be the expense at 6 mills per square foot ? 
 
 Ans. 5231f +ft. boards ; and cost $32,69cts. 6m.+ 
 
 CASE 6. To measure a sector of a circle. 
 
 Definition. A sector is a part of a circle, contained 
 between an arch line and two radii, or semidiameters of 
 the circle. 
 
 RULE. Find the length of the arch by saying, as 160 
 degrees are to the number of degrees in the arch, so is 
 the radius, multiplied by 3,1416, to the length of the arch, 
 which length, divided by 2, and multiplied by the radius^ 
 will become the required area. 
 
OF SUPERFICIES. 200 
 
 EXAMPLE. 
 
 What is the area of the sector of a circle whose radius 
 ia 25 feet, its arch containing 125 ?* 
 As 180: 125 :: 25x3,1416 : 54,5416+ft. length of 
 arch. 54,5416 
 
 Then, x25=681,77ft. Ans. 
 
 2 
 
 RULE 2. Find the area of a circle having the same 
 radius ; then say, as 360 degrees, [the number of de- 
 grees into which all circles are divided,] are to the area 
 of the said circle, so is the number of degrees in the arch 
 of the sector, to the area required. 
 EXAMPLE. 
 
 Required the area of a sector of a circle whose arch 
 contains 65 degrees, and radius 35 feet. Ans. 695^8 q. ft* 
 CASE 7. To jind the area of a segment of a circle. 
 
 Definition. A segment of a circle is any part of a 
 circle cut off by a right line drawn across the circle, 
 which does not pass through the centre, and is always 
 greater or less than a semicircle. 
 
 RULE. Find the area of the sector having the same 
 arch as the segment, by Case 6 ; find also the area of the 
 triangle formed by the chord of the segment and the radii 
 of the sector, by Case 3 ; subtract the area of the latter 
 from that of the former, and the remainder will be the 
 area of the segment, when the segment is less than a 
 semicircle : but the sum af the two areas is the answer* 
 when it is greater, 
 
 EXAMPLE. 
 
 What is the area of the segment, whose arch contains- 
 55 ; its chord 12,5 rods; the perpendicular of its trian- 
 gle 16 rods ; and its semidiameter 17,2 rods-? 
 First, find the area of a circle whose diameter is 34,4 rods 
 As 7 : 22 : : 34,4 : 108,1 + rods circumference*. 
 108,1 34,4 
 
 X =929,66 area of the circle* 
 
 2 2 
 
 Then, as 360 : 929,66: : 55 : 142 + area of the sector, 
 
 * As we have not been able to obtain engravings to represent 
 any of the figures in the preceding or subsequent Examples,, the- 
 Teacher will, we trust, be so good as to draw them, for the PapiS, 
 >u paper or a slate* 
 
 sa 
 
210 MENSURATION* 
 
 And the chord = 12,5 : the perpend. 16. 
 12,5x16 
 > =100 rods, area of the triangle. 
 
 2 
 
 Area of the sector=142 
 
 Area of the triangle = 100 
 
 Area of the segment=42 rods. Ans. 
 NOTE. A regular polygon, whose sides and angles are 
 all equal, may be measured by dividing it into triangles, 
 finding the area of one, and multiplying this area by the 
 number of triangles contained in the polygon. 
 
 CASE 8. To describe, and Jind the areajsf an tllipse or 
 
 oval. 
 
 RULE. To describe an ellipse or aval, draw aline>set 
 one foot of the dividers on the line, as a centre,, and de- 
 scribe a circle ; move the dividers to some other point on 
 the same line, [but not so far but that the dividers in 
 forming a second circle may extend within the firsthand 
 describe a second circle of the same radius as the former ; 
 then, in the two points where the circles intersect, set the 
 dividers to complete the sides of the oval ; and through 
 these intersecting points draw the line called the conju- 
 gate diameter^ crossing the line first drawn called the 
 transverse diameter, in the centre of the oval. 
 
 RULE. To find the area of an ellipse, multiply the 
 transverse, or longest diameter, by the conjugate, or 
 shortest diameter, and their product by ,7854 : and the 
 last product is the area required* 
 
 EXAMPLE. 
 
 If the transverse diameter of aa oval fish pond be 34 
 iods> and the conjugate diameter be 24 rods, what is it* 
 area t 34 x24 X >7854 =640,8864 rods. Ans. 
 
 CASE 9* To Jind the area of a globe or sphere. 
 
 Definition. A sphere or globe is a round solid body, in 
 the middle or centre of which is an imaginary point r 
 from whLeh every part of the surface is equally distant. 
 An apple, or a ball used by children in some of their 
 pastimes, may be called a sphere or globe. 
 
 RuLE.-^-Multiply the circumference by the diameter^ 
 and the product will be the area,, or surface* 
 
OF SOLIDS. 211 
 
 EXAMPLES. 
 
 1. What is the superficial content of the earth, if it be 
 360 degrees in circumference, and every degree measure 
 69 miles ? 
 
 360x69^=25020 circumf. 355 : 113 : : 25020 : 7964 + 
 
 diameter. 
 And 25020x7964 = 199259280 area in squa. miles. Ans. 
 
 2. If the moon's diameter be 2180 miles, what is her 
 area ? Ans. 14928640+ square miles. 
 
 SECTION II. OF SOLIDS. 
 
 Solids are measured by the solid inch, foot, or yard, &c 
 1728 of these inches, that is 12x12x12, make 1 cubic 
 or solid foot. 
 
 CASE I. To measure a Cube. 
 
 Definition. A cube is a solid of six equal sides, each 
 of which is an exact square. 
 
 RULE. Multiply the side by itself, and that product 
 by the same side, and this last product will be the solid 
 content of the cube. 
 
 EXAMPLES. 
 
 1. If the side of the cubic block be 18 inches, or 1 foot 
 and 6 inches, how many solid feet does it contain? 
 
 1ft. 6in. = 1,5ft. and 1,5x1,5x1,5=3,375 solid ft. Ans. 
 
 in. in. in.. 
 Or, 18x18x18 
 
 =3,375 as before. 
 
 1728 
 
 2. Suppose a cellar is to be dug which shall contain 1 
 feet every way, in length, breadth, and depth ; how many 
 solid feet of earth must be taken out to complete it. 
 
 Ans. 1728 sol. ft. 
 
 CASE 2. To find the content of any regular solid, of 
 three dimensions, length, breadth, and thickness, suck 
 as a piece of square timber, urhose length is more than 
 its breadth and depth. 
 
 RULE. Multiply the breadth by the depth or thick- 
 ness, and that product by the length ; the last product i* 
 the solid content. 
 
212 MENSURATION 
 
 EXAMPLES. 
 
 1. How nany solid feet are there in a piece of square 
 timber that is 1 foot and 6 inches, or ISiuches Abroad, 9 
 inches thick, and 9 feet, or 108 inches long ? 
 
 1ft. Gin. = 1,5 foot. ,75x1,5x9=10,125 sol. ft. Ans. 
 9 inches =,75 foot, 
 in. in. in. 
 Or, 18x9x108 
 
 =10,125 as before. 
 
 1728 
 
 In measuring timber, however, you may multiply the 
 breadth in inches by the depth in inches, and that pro- 
 duct by the length in feet : divide this last product by 144, 
 and the quotient will be the solid content in feet, &c. 
 
 2. How many solid feet does a piece of square timber, 
 or a block of marble, contain, if it be 16 inches broad, 
 11 inches thick, and 20 feet long ? 
 
 16x11x20=3520, and 3520 144 =24,4+ sol. ft. Ans. 
 
 3. If a stick of square timber be 15 inches broad, 8 
 inches thick, and 25 feet long, how many solid feet are 
 in it? Ans. 20,8 + feet. 
 
 CASE 3. When the breadth and thickness of a piece of 
 square timber arc given in inches, to Jind how much in 
 length will make a solid foot. 
 RULE. Divide 1728 by the product of the breadth and 
 
 depth, and the quotient will be the length, making a solid 
 
 foot. 
 
 EXAMPLES. 
 
 1. In a piece of square timber 11 inches broad and 8 
 inches deep, what length will make a solid foot 1 
 
 Ilx8=88)l728(19,6+inches. Ans. 
 
 2. In a piece of square timber 18 inches broad and 14 
 inches deep, what length will make a solid foot ? 
 
 Ans. 6,8+inches. 
 
 CASE 4. To measure a cylinder. 
 
 Definition. A cylinder is a round body whose bases or 
 ends are circles, like a round column or stick of timber, 
 of equal bigness from end to end. 
 
 RULE. Multiply the square of the diameter of the 
 base or end by ,7854, which will give the area of the 
 base ; then multiply the area of the base by the length,, 
 and the product will be the solid content* 
 
OF SOLIDS. 213 
 
 EXAMPLES. 
 
 1. What is the solid content of a round stick of timber, 
 or a marble column, of equal bigness from end to end, 
 whose diameter is 18 inches, and length 20 feet ? 
 
 1 8 inches = 1,5ft. l,5x 1,5 x, 7854 = 1,767 15 area of the 
 base. 1,76715x20 length =35,343 solid feet. Ans. 
 Or, 18 x 18 x, 7854 =254,4696 inches, area of the base. 
 
 254,4696x20 
 
 =35,347 as'before. 
 
 144 
 
 2. What is the solid content of a round stick of tim- 
 ber, of equal bigness from end to end, whose diameter is 
 35 inches, and length 35 feet 1 Ans. 233,842 feet. 
 
 CASE 5. Tojind how many solid feet a round stick of 
 timber, equally thick from end to end, will contain, when 
 hewn square. 
 
 RULE. Multiply twice the square of its semidiameter, 
 in inches, by the length in feet ; then divide the product 
 by 144, and the quotient will be the answer. 
 
 EXAMPLES. 
 
 1. If the diameter of a round stick of timber be 22 
 inches, and its length 20 feet, how many solid feet will it 
 contain when hewn square ? 
 
 11x11x2x20 
 
 Half diameter =11, and ==33,6-f ft. the 
 
 144 
 solidity when hewn square, the answer. 
 
 2. If the diameter of a round stick of timber be 24 
 inches from end to end, and its length 20 feet, how many 
 solid feet will it contain, when hewn square, and what 
 will be the content of the slabs which reduce it to a 
 square ? * j 40 feet solidity when hewn square, 
 
 s< \ arid 22,832ft. the solidity of the slabs, 
 
 CASE 6. To Jind how many feet of square edged boards^ 
 of a given thickness, can be sawn from a log of a giv- 
 en diameter. 
 
 RULE. Find the solid content of the log, when made 
 square, by the last Case ; then say, as the thickness of 
 the board, including the saw calf, is to the solid feet, so 
 are 12 inches to the number of feet of boards. 
 
 * -W 
 
214 MENSURATION 
 
 EXAMPLES. 
 
 1. How many feet of square edged boards, l inch 
 thick, including the saw calf, can be sawn from a log 20 
 feet long, and 24 inches diameter 7 
 
 12x12x2x20 
 
 =40ft. solid content when hewn square. 
 
 144 
 
 As 1 : 40 : : 12 : 384 feet. Ans. 
 
 2. How many feet of square edged boards, 1 inch 
 thick, including the saw gap, can be sawn from a log 12 
 feet long, and 18 inches diameter] Ans. 108 feet. 
 
 NOTE. A short rule for finding a number of feet of 
 one inch boards that a log will make, is to deduct of its 
 diameter in inches, and ^ of its length in feet ; then for 
 each inch of diameter that remains, reckon 1 board of 
 the same width as this reduced diameter, and of the same 
 length as this reduced length of the log : thus a log 12 
 feet long, and 12 inches through, gives 9 boards, 9 feet 
 long, 9 inches wide, or 60f feet a log 16 feet long, and 
 16 inches through, gives 12 boards, 12 inches wide, 12 
 feet long, or 144 feet. 
 
 CASE 2. The length, breadth, and depth of any cubical 
 box being given, to Jind how many bushels it will con- 
 tain. 
 
 RULE. Multiply the length, breadth and depth togeth- 
 er, in inches, and divide the last product by 2150,425, the 
 solid inches in the statute bushel, and the quotient will 
 be the answer. 
 
 EXAMPLE. 
 
 There is a square or cubical box ; the length of its bot- 
 tom is 50 inches, breadth of ditto 40 inches, and its depth 
 60 inches ; how many bushels of corn will it hold ? 
 50x40x60 
 
 =55,8+ or 55bush. 3 pecks. Ans. 
 
 2150,425 
 
 CASE 8. To Jind the solidity of a cone or pyramid, 
 
 whether round, square, or triangular. 
 Definition. Solids which decrease gradually from the 
 base till they come to a point, are generally called cones 
 or pyramids, and are of various kinds, according to the 
 
or SOLIDS. 215 
 
 figure of their bases ; round, square, oblong, triangular, 
 &c. ; the point at the top is called the vertex, and a line 
 drawn from the vertex, perpendicular to the base, is call- 
 ed the height of the pyramid. 
 
 RULE. Find the area of the base, whether round 
 square, oblong, or triangular, by some one of the forego- 
 ing rules, as the case may be ; then multiply this area by 
 one-third of the height, and the product will be the solid 
 content of the pyramid. 
 
 EXAMPLES. 
 
 1. What is the solid content of a true-tapered round 
 stick of timber, 24 feet perpendicular length, 15 inches 
 diameter at one end, and a point at the other? 
 
 15xl5x,7S54 X 8 
 
 =9,8 175 solid feet. Ans. 
 
 144 
 
 2. What is the solid content of a square stick of timber 
 of a true taper, 30 feet perpendicular length, 18 inches 
 square at one end, and a point at the other 1 Ans. 22feet. 
 
 3. What is the solid content of a triangular tapering 
 stick of timber, 21 feet long, 10 inches each side of the 
 triangle, 8f inches the perpendicular of the triangle at 
 the large end, and the other end a point ? 
 
 Half perpendicular=4,33 and 4,33x10x7 
 
 =2, 1 ft. + Ans. 
 
 144 
 
 NOTE. If a stick of timber be hewn three square, and 
 be equal from end to end, you find the area of the base as 
 in the last question, in inches, multiply that area by the 
 whole length, and divide the product by 144, to obtain 
 the solid content. 
 
 4. If a stick of timber be hewn three square, be 12 feet 
 long, and each side of the base 10 inches, the perpendic- 
 ular of the base being 8f inches, what is its solidity ? 
 
 Ans. 3,6+ feet. 
 
 CASED. To Jind the solidity of the frustum of a eone 
 
 or pyramid. 
 
 Definition. The frustum of a cone is what remains 
 after the top is cut off by a plane parallel to the base, and 
 is in the form of a log greater at one end than theother a 
 whether round, or hewn three or four square, &c. 
 
216 MENSURATION 
 
 RULE. If it be the frustum of a square pyramid, mul- 
 tiply the side of the greater base by the side of the less ; to 
 this product add one third of the square of the difference 
 of the sides, and the sum will be the mean area between 
 the bases ; then multiply this sum by the height, and it 
 will give the content of the frustum. Or, if it be a taper- 
 ing square stick of timber, take the girth of it in the mid- 
 dle ; square of the girth, (or multiply it by itself in inch- 
 es ;) then say, as 144 inches to that product, so is the 
 length, taken in feet, to the content in feet. 
 
 EXAMPLE. 
 
 What is the content of a tapering square stick of tim- 
 ber, whose side of the largest end is 12 inches, of the 
 least end, 8 inches, and whose length is thirty feet, calcu- 
 lating it by both rules ? 4x4 
 
 By the first Rule : 12x8=*96. 128=4 =5 
 
 _ 3 
 
 And 96+5^x30 
 
 =2lft. Ans. 
 
 144 
 By the second Rule : 12+8 
 
 =10in.= of the girth in 
 
 the middle. 2 
 
 Then 10xlO=100=area in the middle of the stick. 
 And, as 144 : 100 : : 30ft. : 20, 83+ feet. Ans. 
 
 RULE. If it be a triangle pyramid, or a tapering 
 three square stick of timber, multiply the sum of the 
 mean area, as found in the first rule, by ,433 and that 
 product by the height or length. Or, multiply the area 
 in the middle, as found in the second rule, by ,433 and 
 then state the proportion as before. 
 EXAMPLE. 
 
 What is the content of a tapering three-square stick of 
 timber, whose side of the largest end is 15 inches, of the 
 least end, 6 inches, and whose length is 40 feet, calcu- 
 lating it by both rules ? 9x9 
 
 By the first Rule: 15x6=90. 156=9. =27. 
 
 3 
 
 And90+27x,433x40 
 
 = 14,0725ft. Ans. 
 
 144 
 
OF SOLIDS. 217 
 
 15+6 
 
 By the second Rule : =10,5 in.= of the girth 
 
 2 
 
 in the middle, if it were four-square. 
 Then 10,5 x 10,5 x, 433 =47,73825in.-=area in middle. 
 And, as 144: 47,73825 : : 40 feet : 13,260625ft. Ans, 
 
 RULE. If it be a circular pyramid or cone, multiply the 
 diameters of the two bases together,and to the product add 
 one third of the square of the difference of the diameters ; 
 then multiply this sum by ,7854 and it will be the mean 
 area between the two bases ; multiply this area by the 
 length of the frustum, and it will give the solid content. 
 
 Or multiply each diameter into itself; multiply one 
 diameter by the other ; multiply the sum of these pro- 
 ducts by the length ; annex two ciphers to the product, 
 and divide it by 382 ; the quotient will be the content, 
 which divide by 144 for feet as in other cases. 
 
 EXAMPLE. 
 
 What is the solid content of a tapering round stick of 
 timber, whose greatest diameter is 13 inches, the least 
 6 inches, and whose length is 24 feet, calculating it by 
 both rules ? 
 
 13x6,5=84,5 136,5=6,5 6,5x6,5 
 
 =14,083+ 
 
 3 
 
 And 84,5+1 4,083 x, 7854x24 
 
 = 12,904 + feet. Ans. 
 
 144 
 By the second Rule ;* 
 
 13x13+6,5x6,5+13x6,5x2400 
 
 =1858,1 15+ 
 
 382 
 And 1858,115~144 = 12,903+ft. Ans. 
 
 * To find the content of timber in the tree, multiply the square 
 of 1-5 of the circumference at the middle of a tree, in inches, by 
 twice the length in feet, and the product divided by 144 will be 
 the content, extremely near the truth. In oak an allowance of 
 1-10 or 1-12 must be made for the bark, if on the tree ; in other 
 wood less* Trees of irregular growth must be measured in parts. 
 T 
 
218 MENSURATION OF SOLIDS. 
 
 CASE 10. To Jind the solid content of a Sphere or Globe. 
 
 NOTE. For definition of a Globe, see Case 9 of Su- 
 perficies. 
 
 RULE. Find the superficial content by Case 9 of Su- 
 perficies ; multiply this surface by one-sixth of the diam- 
 eter, and it will give the solidity. 
 
 Or, multiply the cube of the diameter by ,5236 and 
 the product will be the solidity.* 
 
 EXAMPLE. 
 
 What is the solidity of our earth, if its diameter be 
 7957 f miles, nearly, and its circumference at the equator 
 be just 25000 miles? 
 
 7957,75x25000x7957,75-f-6=~263857106187,5-f solid 
 
 miles. Ans. 
 
 CASE 11. To Jind the solid content of a frustum or seg- 
 ment of a Globe. 
 
 Definition. The frustum of a globe is any part cut 
 off by a plane. 
 
 RULE. To three times the square of the semidiameter 
 of the base, add the square of the height ; multiply this 
 sum by the height, and the product again by ,5236; the 
 last product will be the solid content. 
 
 EXAMPLE. 
 
 If the height of a coal-pit, at the chimney, be 9 feet, 
 and the diameter at the bottom be 24 feet, how many 
 cords of wood does it contain, allowing nothing for the 
 chimney ? 
 
 24-r-2=12=semidiam. 12x12x3=432. 9x9=81. 
 
 And 432+81 x9x,5236 
 
 =18,886+ cords. Ans. 
 
 128=solid feet in a cord. 
 
 *If the diameter of a sphere be 1, its solidity will be ,5236; 
 and if its circumference be 1, its solidity will be ,016887. 
 
GAUGING. 219 
 
 SECTION III. 
 
 OF CASK GAUGING. 
 
 Definition. Gauging is the finding of the content of 
 any Cask, Box, Tub, or other Vessel. 
 
 Among the many different rules for gauging, the fol- 
 lowing is as exact as any. 
 
 RULE. Take the diameter at the hung and head, and 
 length of the cask ; subtract the head diameter from the 
 bung diameter, and note the "difference. 
 
 If the staves of the cask be much curved or bulging 
 between the bung and head, multiply the difference of di- 
 ameters by ,7 ; if not quite so much curved, by ,65 ; if 
 they bulge yet less, by ,6 ; and if they are almost or quite 
 straight, by ,55 and add the product to the head diame- 
 ter ; the sum will be a mean diameter. 
 
 Square the mean diameter, thus fou-nd, and multiply 
 the square by the length ; divide the product by 359 for 
 ale or beer gallons, and by 294 for wine gallons. _ 
 
 NOTE. 1. To measure the length of the cask, take 
 the length of the stave ; then take the depth of the chimes, 
 which, with the thickness of the heads, (that are 1 inch, 
 lvj- inch, or 2 inches, according to the size of the cask) 
 being subtracted from the length of the stave, leaves the 
 length within. 
 
 2. In taking the bung diameter observe by moving the 
 rod backward and forward whether the stave opposite to 
 the bung, be thicker or thiner than the rest, and if it 
 be, make allowance accordingly. 
 EXAMPLE. 
 
 How many ale and wine gallons will a cask contain, 
 whose bung diameter is 30 inches, head diameter 25 in- 
 ches and length 40 inches ? 
 
 3025=5. 5x,7=3,5 25 + 3,5=28,5 mean 
 
 diam. 
 
 28,5x28,5x40 
 
 ..-=90,5+ ale gal. Ans. 
 
 359 
 
 28,5x28,5x40 
 
 =110,51 +wine gal. Ans. 
 
 294 
 
220 GAUGING. 
 
 Or, by the sliding rule. On D. is 18,94 the gauge- 
 point for ale or beer gallons, marked A. G : and 17,14 
 the guage-point for wine galons, marked W. G. Set the 
 gauge-point to the length of the cask on C. and against 
 the mean diameter, on D. you will have the answer in ale 
 or wine gallons, accordingly as which gauge-point you 
 make use of.* 
 
 CASE 2. To gauge round tubs, fyc. 
 RULE. Multiply one diameter by the other, and to 
 that product add one third of the square of their differ- 
 ence ; multiply this sum by the length, and divide as be- 
 fore for beer or wine. 
 
 EXAMPLES. 
 
 What is the content, in beer and wine gallons, of a 
 round tub, whose diameter at the top, within, is 40 inches, 
 and at the bottom 34 inches, and the perpendicular height 
 
 36 inches ? 
 
 6x6 34x40+12x36 
 
 =12. =137+beer gal. 
 
 40 4=6 3 359 Ans. 
 
 And 34x40+ 12x36 
 
 =168 wine gal. Ans. 
 
 294 
 
 CASE 3. To gauge a square vessel. 
 RULE. Multiply the length by the breadth, and that 
 product by the depth ; then divide by 282 for beer or ale, 
 (the inches in a beer or ale gallon,)and by 231 for wine, 
 &c., (the inches contained in a wine gallon,) and the 
 quotient will be the answer. 
 
 * A rule which has been given as generally more exact, is this 
 multiply the product of the square of the mean diameter and the 
 length, by 34, and point off four places from the right of the pro- 
 duct ; the figures on the left of the point will be the gallons, and 
 those on the right decimal parts of a gallon, in wine or cider* 
 Let the dimension be taken exactly in inches and tenths. Take 
 the preceding- Example in Case 1. 
 
 30+25 
 
 =27,5 mean diam. 
 
 2 
 
 and 27,5x27,5x40x34= 1 028500,00 =102^gall on s of 
 wine or cider ^ which i&7y% 6 <jgal. less than by the other.. 
 
SHIPb' TONNAGE, &C. 221 
 
 EXAMPLE. 
 
 If a square vessel he 80 inches in length, 60 in breadth, 
 and 40 inches deep, what is its content in beer and wine 
 gallons ? 
 80x60x40 wine gal. 80x60x40 beer gal. 
 
 . -63l,l(3+Ans. =680,85+ Ans, 
 
 231 282 
 
 NOTE. The content of any vessel, in feet, gallons, 
 and bushels, may be thus found : Measure the inside of 
 the vessel, according to the rule of the figure, and find 
 the content in cubic inches ; then, 
 
 C 1728 } and the C Cubic feet. 
 
 TV i i j 282 f quotient y Ale or beer gal. 
 
 5 b ^ ) 231 f will be the ^ Wine gallons. 
 
 \. 2150,425 ) content in V. Bushels. 
 
 SECTION IV. To find the tonnage of a ship. 
 
 RULE. Multiply the length of the keel by the breadth 
 
 of the beam, and that product by the depth of the hold ; 
 
 divide the last product by 95, arid the quotient is the 
 
 tonnage. If double decked, half the breadth is the depth. 
 
 EXAMPLE. 
 
 If a ship be 72 feet by the keel, 24 feet by the beam, 
 and 12 feet deep, what is the tonnage 1 
 
 Ans. 
 
 Or, RULE 2. Multiply the length of the keel by the 
 breadth of the beam, and that product again by half the 
 breadth of, the beam ; divide the last product by 94, and 
 the quotient is the tonnage.* 
 
 EXAMPLE. 
 
 What is the tonnage of a ship that is 84 feet by the 
 keel, arid 28 feet by the beam ? 23-f-2 = 14 
 
 84x28x1 4^94=350,29 -f tons. Ans, 
 NOTE. The breadth of the beam added to two thirds 
 the length of the keel, gives the length of a ship's main-* 
 
 * Rule established by Congress. For double decked vessels ; length 
 frora fore part of main stem to ~afterpart of stern post, above tipper deck ; 
 breadth at videst part, above main wales, La.f of which is called the depth .;, 
 deduct from length 3-5ths of breadth 3 multiply the remainder 
 
222 STRENGTH OF CABLES. SHIPS* BURTHEN, &C. 
 
 mast : Therefore the length of the mainmast of the ship 
 last mentioned, is 84x2-f-3+28=84 feet. 
 
 To find the thickness, the proportion is, as 84 to 28, so 
 is the length of a mast in feet, to its thickness in inches. 
 Consequently the thickness of the mast whose length was 
 just found, is 28 inches. 
 
 SECTION V* TfrfindtJie weight of anchors which cables 
 may sustain. 
 
 RULE. As the strength of cables, and consequently 
 the weights of their anchors, are proportioned to the 
 cubes of their peripheries ; therefore, as the cube of the 
 periphery of any cable, is to the weight of its anchor, so 
 is the cube of the periphery of any other cable, to the 
 weight of its anchor. 
 
 EXAMPLES. 
 
 1. If a cable of 6 inches round require an anchor of 
 2cwt. what would be the weight of an anchor, for a 12 
 inch cable ? cwt. 
 
 6x6x6 : 2,25 :: 12x12x12 : 18cwt. Ans. 
 
 2. If a 12 inch cable required an anchor of 18cwt. what 
 must the circumference of a cable be, for an anchor of 
 2cwt. T cwt> cwt. 
 
 18 ; 12x12x12 : : 2,25: 216. ^216=6in. Ans. 
 
 SECTION VI. From one solid's capacity, to find another's. 
 
 RULE. As the cube of any dimension is to its given 
 
 weight* so is the cube of any like dimensions to its weight. 
 
 EXAMPLES. 
 
 1. If a ship of 300 tons' burthen be 75 feet by the 
 keel, what is the burthen of one,. 100 feet by the keel, of 
 tike fornv, 25|fc. a qiv ? 
 
 tons. tons. cwt. \fo* 
 
 75 3 : 300 i : 100 3 s 71 1 2 22f . Ans. 
 
 2. If brass cannon, lljinch. diameter, weigh lOOOffe. 
 what will another, 208& inch diameter,, of like metal 
 and shape, weigh * Ans. 5942,5697ft. + 
 
 and the product by depth ; divide by 95; the quotient is the true tonnage. 
 For single decked, take depth from under side of deck plauk to ceiling ia 
 t&e bolci;, and proceed as before^ 
 
WOOD MEASURE. - ASSESSMENT OF TAXES. 223 
 
 SECTION VII. 
 OF WOOD AND BARK MEASURE. 
 
 CASE 1. To find the solid content of wood 4* bark. 
 RULE. Multiply the length, breadth, and thickness to- 
 gether, agreeably to the rule of Duodecimals, and the last 
 product will be the solid content of the pile, parcel, or load. 
 
 EXAMPLE. 
 
 If a load of wood be 8 feet 4 inches long, 3 feet 8 in- 
 ches wide, and 4 feet 6 inches high, how many cubic feet 
 does it contain ? ft. / ft. y ft. t 
 
 8 4x3 8x4 6=137 sol. ft. Ans. 
 CASE 2. To Jind how many cords of wood or bark are 
 
 contained in any pile, 4*c. 
 
 RULE. Find the solid content as before, and divide 
 that product by 128 ; the quotient will be the cords, and 
 the remainder cubic feet, or so many 128ths of a cord. 
 Or, divide the solid content of the pile, &c. by 16, and 
 the quotient will be cord-feet, 8 of them being 1 cord, 
 and the remainder so many 16ths of a cord-foot. 
 EXAMPLES. 
 
 1. In a pile or load of wood 9 feet 4 inches long, 3 
 feet 8 inches wide, and 4 feet 9 inches high, how many 
 cords, and how many cord-feet ? 
 
 ft-, ft- , ft- < *olft. iM ft. 
 
 94x38x49=16268 And 162 6 8-f- 128=1 cord, 
 
 and 34 sol. ft. 6 8. Ans. 
 
 Or, 162 68-4-16=10|cord ft. 6 8=1 cord 2J- feet. Ans. 
 
 2. If a load of wood or bark be 8 feet long, 4 feet wide, 
 and 2 feet 6 inches high, how many cord-feet does it con* 
 tain ? Ans. 5 feet, or of a cord. 
 
 MODE OF ASSESSING TAXES. 
 
 It may not perhaps, be here amiss to show the general method of 
 Assessing- Taxes. But as the quantity of new matter with 
 which we have enlarged this edition of our Work, has extend- 
 ed its pages considerably beyond the limits first intended, a 
 brief explanation of the general principle and rule, will, we 
 trust* fully suffice far the purpose. 
 
224 ASSESSMENT OF TAXES. 
 
 ARGUMENT. 
 
 There is a certain town which contains 8 inhabitants, whom we will call 
 A, B, C, D. E, F, G, and H. The town is divided into 2 school districts or 
 classes, which are numbered 1 and 2. A, B, C, and D, form District INo. 1, 
 and E, F, G, and H, No. 2. On these inhabitants the following 1 taxes are to 
 be assessed, namely : 
 
 State, $14, 88cts. 6m. 
 
 County, 19, 84cts. 8m. 
 
 Town, 39, 69cts. 6m. 
 
 School, 29, 77cls. 2m. 
 
 fixed ; and for that purpose all assessors consult, of course, the latest acts that 
 have been passed on the subject. 
 
 Let a poll be taxed $1,30 ; an acre of orchard 25cts. ; an acre of tillage 
 16cts. 5 an acre of mowing IGcts. 5 an acre of pasturage 4-cte. $. an ox of five 
 years 35cts. ; and a cow 20cts. 
 
 In order to find each person's proportion of the several taxes, and each 
 school district's proportion of school money, according to the rateable estates 
 of the members of each district or class, or according to the number of schol- 
 ars in each district ; each man's inventory must be taken, and the amount 
 cast by the following rule. 
 
 RULE. Multiply the value of a poll by his number of polls ; his acres of 
 orchard by the tax- value of one ; his number of oxen by the tax-rate of one j 
 and so of every other kind of property ; add the products, and the sum is the 
 amount of his rateable estate; find the amount of all in the same way ; add 
 these amounts, and their sum is the value of the inventory of the town. 
 
 I demand the rateable estate of A, who has 
 2 Polls, at $1,30 amount to $2,60 
 
 2 Acres of tillage, at 0,16 0,32 
 
 5 Acres of mowing", at 0^16 0,80 
 
 2 Oxen, at 0,35 0,70 
 
 Amount of A*s rateable estate, $4,42 
 
 Find the other amounts, in the following* inventory, in the 
 same way. / 
 
 To prove the Inventory. 
 
 RULE. Add up the column of polls, and multiply the sum by 
 the value of one: add up each of the other columns, and multi- 
 ply its sum by the tax-rate of one in that column; then add the 
 several amounts of the columns tog-ether, and the sum will be 
 equal to the total amount of the fnventory, if the work be rig-lit. 
 
 EXAMPLE. 
 
 The total amount of rateable estates in the following* invento- 
 ry, is $49, 62cts. 
 
 " And proceeding by the method given in the rule of proof, the 
 sum of the products is $49, 62cts, It is, therefore, evident the 
 work is right. 
 
 * Assess money taxes so far over the sum to be raised, as to meet abate- 
 ments. 
 
ASSESSMENT OF TAXES. 
 
 225 
 
 
 
 
 
 1 
 
 c 
 
 
 
 3 
 
 a 
 "o 
 
 
 2 
 
 3 
 
 NOTE. If any teacher think it 
 
 
 
 
 
 o 
 
 "S 
 
 P 
 
 
 
 best to proportion the School Money 
 
 
 
 O ^ 
 
 P 
 
 
 
 as 
 
 &H 
 
 s-i 
 
 
 <$ W 
 
 between the districts, according to the 
 
 GO 
 
 
 d 
 
 
 
 
 iO 
 
 
 ^5 
 
 number of scholars in each, instead of 
 
 Q 
 
 CO 
 
 c .a 
 
 Q O 
 
 
 
 o 
 
 "o 
 
 G 
 
 
 S 
 
 -*-i fw 
 
 by the value of rateable estates in 
 
 rt 
 
 
 B 
 
 y 
 
 6 
 
 Q 
 
 Q 
 
 X 
 
 > 
 
 Q 
 
 o 
 
 each, let the scholar do it so ; and let 
 
 
 
 
 
 < 
 
 
 
 O 
 
 
 o 
 
 district No. 1 contain 15, and district 
 
 A 
 
 B 
 
 2 
 
 1 
 
 3 
 
 2 
 
 5 
 
 5 
 2 
 
 4 
 
 2 
 
 1 
 
 $4,42 
 3,53 
 
 No. 2, 20 scholars. The inventory 
 here given, though it exhibits but a 
 few rateable articles, will serve to 
 
 C 
 D 
 
 4 
 I 
 
 6 
 
 3 
 
 8 
 
 10 
 
 2 
 
 2 
 1 
 
 9,96 
 1,50 
 
 explain the principle. As minors now 
 pay no poll-tax in Maine, no person 
 
 E 
 
 3 
 
 
 10 
 
 5 
 
 8 
 
 
 2 
 
 7,02 
 
 can properly, have more than 1 poll ; 
 
 F 1 
 
 5 
 
 4 6 
 
 5 
 
 2 
 
 1 
 
 5,25 
 
 though he may pay the tax for his 
 
 G 
 
 2 
 
 4 
 
 8 
 
 4 
 
 6 
 
 2 
 
 
 6,46 
 
 workmen, and his sons who are of 
 
 H 
 
 5 
 
 
 12 
 
 8 
 
 12 
 
 2 
 
 3 
 
 11,48 
 
 age. 
 
 Total Amt. $49,62 
 
 To find each person's proportion of any tax. 
 
 RULE. Say, as the total amount of the Inventory of the 
 town, is to the sum to be raised in each tax, so is 1 dollar to that 
 part of the tax which one dollar of the Inventory, or rateable 
 estate, must pay : then, taking- the same numbers for the first 
 and second terms, and one cent for the third term, of a new 
 stating-, find what part of the tax one cent of the Inventory, or 
 rateable estate, must pay ; and from these two operations form 
 two tax tables ; one for dollars, from 1 dollar to 11, or farther, if 
 deemed necessary ; and the other for cents, from one cent to 
 90. Then by means of these two tables, make out each per- 
 son's tax. 
 
 1. To make the State tax, the sum to be raised being* $14,88cts. 
 6m., and the total amount of the foregoing- Inventory $49,62cts. 
 
 As $49,62cts. : $14, 88cts. 6m. : : $1, OOcts. : $0, 30cts. 
 
 And as $49, 62cts. : $14, 88cts. 6m. : : ,01ct. : ,00cts. 3m. 
 
 Therefore, $1 of the Inventory pays 30 cents ; and 1 cent of 
 the Inventory pays 3 mills ; by which make the following- two 
 Tables. 
 
 DOLLAR TABLE, 
 
 from $1 to $11. 
 
 $ $ cts. 
 
 I pays 0, 30 
 
 CENT TABLE, 
 
 from 1 Cent to 90 Cents. 
 
 2 
 
 3 
 
 4 
 
 5 
 
 6 
 
 7 
 
 8 
 
 9 
 
 10 
 
 11 
 
 0, 60 
 
 0, 90 
 
 1, 20 
 1, 50 
 
 1, 80 
 
 2, 10 
 2, 40 
 
 2, 70 
 
 3, 00 
 3, 30 
 
 cts. 
 
 cts. 
 
 m. 
 
 cts. 
 
 cts. 
 
 1 pays 
 
 
 
 3 
 
 30 pays 
 
 9 
 
 2 
 
 
 
 6 
 
 40 
 
 12 
 
 3 
 
 
 
 9 
 
 50 
 
 15 
 
 4 
 
 1 
 
 2 
 
 60 
 
 18 
 
 5 
 
 1 
 
 5 ' 70 
 
 21 
 
 6 
 
 1 
 
 8 ' 80 
 
 24 
 
 7 
 
 2 
 
 1 
 
 90 
 
 27 
 
 8 
 
 2 
 
 4 
 
 
 
 9 
 
 2 
 
 7 
 
 
 
 10 
 
 3 
 
 
 
 
 20 
 
 6 
 
 i 
 
 
226 BOOK-KEEPING. 
 
 The tax is now to be made on each rateable estate, as it stands 
 in the Inventory, by means of these tables. First, What js A's 
 tax, whose rateable estate is $4. 42cts. ? 
 By the table, $4 pay $1, 20cts. 
 
 40cts pay 0,12 
 and 2 " ^ 0,00, 6m. 
 
 Amount $1, 32, 6 A's tax. 
 
 Or, having- found what part of the tax one cent of the Invento- 
 ry will pay, you may, instead of making- tables, multiply the 
 number of cents, in each person's Inventory, by what one cent 
 pays and the product will be his tax. 
 Now, to find A's tax by this method : 
 One cent pays 3 mills, or ,3 of a cent. 
 Therefore, 442cts. 
 ,3 
 
 132,6=132 T V5ts. or $1, 32cts. 6m. as before. 
 
 Find by these methods, the State tax of all the other persons. 
 Then, to know if your work be rig-lit, add the several persons' 
 taxes tog-ether, 'and see if the sum be just equal to the $14 88cts. 
 6m. that was raised for the State, which it must be, because the 
 proportion is even. 
 
 Next, find each person's County tax in like manner, taking new 
 stating-s, and forming- new tables : and thus proceed with each 
 particular tax, till you have gone through the whole, proving 
 each part as before noted. 
 
 Lastly, form your tax list, setting- down the names therein al- 
 phabetically, and carrying- out in a line from each the separate 
 sums of the respective taxes, together with the total amount of 
 each. When done, give them a general proof, by adding togeth- 
 er the several sums that were to be raised for State, County, &c. 
 taxes : and then the total amounts of each person's taxes; which 
 two sums will come exactly alike, if there be no errour, in any 
 part of the work. 
 
 BOOK-KEEPING. 
 
 DIRECTIONS FOR THE LEARNER. 
 
 Having- ruled your books in the proper form, copy inlo the Daybook one 
 day's accounts; then calculate them upon your slate or waste-paper, to 
 find if they be rightly cast up, and to exercise you in calculations. Next, 
 rule your slate or waste-paper in the form of the Leger, and upon it post 
 the accounts that you have copied in the Daybook, with their date prefix- 
 ed 5 observing to set on the Dr. side of each person's account, those ac- 
 counts to which he is Dr. in the Daybook, and on the O. side of his ac- 
 count, those by which he is Cr. And if any account consist of but one 
 article, you are to express it particularly with its amount, in the columns; 
 but if it consits ; of several articles, write To or By Sundries, placing the 
 
 JI^L 
 
BOOK-KEEPING. 
 
 227 
 
 sum of the amounts of all the articles in the columns. After the accounts 
 are, by correcting if necessary, placed according to the teacher's mind, 
 transcribe them into your Leger, leaving a proper space, under each per- 
 son's name, to receive more accounts. Then under the proper letters in 
 the Alphabet, enter those names with the pages where they stand in the 
 Leger 5 and, lastly, write the Daybook pages to the several accounts in 
 the Leger, by which you can readily refer to the page of the Daybook on 
 which any Leger entry may be found, making at the same time, the marks 
 on the Daybook which denote the several accounts to be posted. Do the 
 same with the next day's accounts : and so on till the whole be finished. 
 But observe that you must not enter any person's name down again which 
 has been entered before, till the space first assigned to it shall be filled with 
 articles ; and then the account must be tranferred to a new place, as you 
 may observe is done with George Sampson's account. 
 EXAMPLE. 
 
 Suppose David Davis owes me 450 dollars for the balance of an account 
 with him, April 1st. 1822; the next day, April 2d. I buy of him 200 bush- 
 els of wheat at 1 dollar 50 cents per bushel, and 100 bushels of corn at 75 
 cents per bushel ; the next day, April 3d, I sell Jonathan Worth 150 bush- 
 els of wheat' at 1 dollar 75 cents per bushel j April 4th, Jonathan Worth 
 pays me 200 dollars in cash, and David Davis pays me 50 dollars in cash : 
 'required the Daybook and Leger of the transaction. 
 
 DAYBOOK, NO. 1. 
 
 Hallowdl, April 1, 1822. 
 
 = 
 
 David Davis. 
 To balance due on old account, 
 /?**'*> 
 
 Dr. 
 
 $ 
 450 
 
 cts 
 
 00 
 00 
 
 00 
 50 
 00 
 
 = 
 
 David Davis, 
 By 20 bushels wheat, . 
 100 do. corn, 
 
 April 3 
 
 Or. 
 
 at $1, 50 
 75 
 
 300 
 75 
 
 375 
 
 
 
 Jonathan Worth, 
 To 150 bushels wheat, 
 
 /?rT*i/ A. 
 
 Dr. 
 
 at $1,75 
 
 Cr. 
 
 262 
 
 200 
 
 Jonathan Worth, 
 By cash in part for wheat, 
 
 David Davis, 
 By cash, fifty dollars, 
 
 Cr. 
 
 50 
 
 00 
 
 To post the above accounts, open an account for David Davis, debit him 
 for 450 dollars; and for the second day's transaction credit him for 375 
 dollars : for the third open an account for Jonathan Worth, debiting him 
 for 262 dollars 50 cents ; and for the fourth day credit him for 200 dollars, 
 and credit David Davis for 50 dollars. 
 
228 
 
 L EGER NO. 1. 
 1822. Dr. David Davis, 
 
 April 1 
 April 4 
 
 To balance of old account, 
 To balance of above account, 
 
 227 
 
 $| 
 450 
 
 450 
 
 Cts. 
 00 
 
 00 
 
 25 
 
 00 
 
 1822. 
 
 Dr. Jonathan Worth, 
 
 
 
 
 April 3 
 
 i 
 
 To wheat, 
 
 227 
 
 262 
 
 50 
 
 262 
 
 50 
 
 4 
 
 To balance of old account, 
 
 62 
 
 50 
 
 By the above Leger it appears that the balances are in 
 my favour, which, added to the cash I have on hand, and 
 the goods unsold, show the amount of my stock, which 
 compared with my original stock, will show my profit or 
 
 loss, viz. 
 
 $. Cts. 
 25 00 
 62 50 
 25000 
 
 David Davis owes me 
 Jonathan worth do. 
 I have in cash, 
 Wheat unsold 50 bushels, 
 
 valued at prime cost, $1,50 
 Corn do. 100 bushels do. at 75 cts. 75 00 
 
 7500 
 
 Amount of my stock, 
 My original stock wai 
 
 I have therefore gained 
 
 487 50 
 450 00 
 
 37 50 
 
(1*) 
 
 LEGER, NO. I. 
 
 S29 
 
 1822. Contra 
 
 Cr. 
 
 April 2 
 4 
 
 By Sundries, - 
 By Cash, 
 By balance to new account, 
 
 227 
 
 $ 
 375 
 50 
 25 
 
 Cts. 
 00 
 00 
 00 
 
 00 
 
 450 
 
 
 
 1822. 
 
 April 4 
 
 Contra Cr. 
 
 
 
 
 By cash, - - 
 By balance to new account, 
 
 227 
 
 200 
 62 
 
 00 
 50 
 
 50 
 
 262 
 
 
 
 NOTE. If you should enter any thing in your Leger 
 under a wrong title, or in any other way false, it should 
 not be blotted out, but marked thus (x) in the margin 
 against it ; and write on the opposite side Err our per con- 
 tra, with the sum against it, and make the same mark in 
 the margin. 
 
 * Opposite pages in the Leger are both numbered alike. 
 
230 
 
 0) 
 DAYBOOK, No. 2. 
 
 Hallotoell, April G, 1827. 
 
 George Simpson 
 To balance due on old account, 
 
 Dr. 
 
 John Barton Dr. 
 
 To balance due on account of 6 ) 
 Hhds. Wine. } 
 
 William Reed 
 
 Cr. 
 
 By balance due him on account of > 
 English goods pr. invoice, ) 
 
 Thomas Tilton Cr. 
 
 By balance due him for 6 months 
 service on farm, 
 
 April 8. 
 
 Charles Prince Dr. 
 
 To 1 2 I7tb- sugar 25fca qr. at $12 
 2 bbls. superfine flour, at 7 50 
 
 1 do. mess pork, 
 
 Richard Lewis Dr. 
 
 To 2yds. superfine broadcloth at $6 
 4 - cassimere, best blk. 15s. 
 
 1^ doz. buttons, 5s. 6d. 
 
 1 do. small do, 
 4 skeins silk, 
 4 sticks twist. 
 
 By cash in part, 
 
 Cr. 
 
 200 
 
 cts. 
 
 340 
 
 462 
 
 44 
 
 90 
 
 20 
 15 
 25 
 
 10 
 00 
 00 
 
 60 
 
 10 
 
 12 
 
 10 
 
 1 
 
 00 
 00 
 
 37 
 38 
 
 25 
 
 24 
 15 
 
 25 
 00 
 
(2) 
 Hallowell, April IS, 1827. 
 
 231 
 
 = 
 
 Thomas Tilton. 
 To cash in part of the balance due 
 his order paid Samuel Lane, 
 
 A nr jl 14 
 
 Dr. 
 
 him, 
 
 $ 
 50 
 21 
 
 cts. 
 00 
 50 
 
 71 
 
 50 
 
 George Simpson 
 By cash (per rect.) 
 
 Anril 4 , . 
 
 Cr. 
 
 41 
 
 00 
 
 Richard Lewis Dr. 
 To 5 pieces India cottons, at 26s, 
 7 yds. cotton cambrick, 4s. 6d. 
 6 " col'd do. 3s. 
 
 Anril rtc i 
 
 21 
 5 
 3 
 
 67 
 25 
 00 
 
 29 
 
 92 
 
 George Simpson 
 By check on Gardiner Bank for > 
 one hundred and fifty dollars, ) 
 
 Anril ^0 
 
 Cr. 
 
 150 
 
 00 
 
 John Barton 
 By cash rec'd for 25bbls. beef sold 
 at $10,50cts. 
 
 , Tnnr f\ L- 
 
 Cr. 
 
 } 
 
 262 
 
 50 
 
 George Simpson 
 To 10 gallons wine, at $ 
 
 Dr. 
 
 1,20 
 
 12 
 
 00 
 
 INDEX TO LEGER NO 2. 
 
 B Barton John, 
 
 L Lewis Richard, 
 
 P Prince Charles, 
 
 R Reed William, 
 
 S Simpson George, 
 
 T Tilton Thomas, 
 
 folio I 
 
 2 
 
 2 
 
 1 
 
 1,2 
 
 a 
 
232 
 
 CO 
 LEGER, No 2. 
 
 1827. Dr. George Simpson, 
 
 April 6 
 
 To balance, 
 
 F 
 
 ol. 
 
 1 
 
 $ 
 200 
 
 Cts. 
 00 
 
 
 Transferred to folio 2. 
 
 
 
 200 
 
 00 
 
 
 
 
 
 
 
 1827. 
 
 Dr. John Barton, (1 
 
 3; 
 
 ith 
 
 ) 
 
 
 April 6 
 
 To balance for wine, 
 
 
 1 
 
 I $ 
 346 
 
 Cts. 
 00 
 
 1827. 
 
 Dr. William Reed, 
 
 
 
 
 
 
 
 
 
 
 
1827. 
 
 (l) 
 LEGER, No. 2. 
 
 Contra 
 
 233, 
 
 Cr. 
 
 April 14 
 25 
 June 1 
 
 By cash, 
 By check on Gardiner Bank, 
 By balance transferred to folio 2, 
 
 Fol 
 2 
 2 
 
 $ 
 41 
 150 
 9 
 
 Cts. 
 00 
 00 
 00 
 
 
 
 , 
 
 j 
 
 200 
 
 00 
 
 1827. 
 
 Contra Cr. 
 
 
 
 
 April 30 
 
 
 
 By cash, 
 
 2 
 
 262 
 
 50 
 
 1827. 
 
 Contra Cr 
 
 
 
 
 April 6 3 
 
 3y balance due, 
 
 1 
 
 462 
 
 44 
 
 v a 
 
234 (2) 
 1827. Dr. Thomas Tilton, 
 
 April 13 
 
 To sundries, 
 
 2 
 
 $ i 
 71 
 
 as. 
 
 50 
 
 1827. Dr. Charles Prince, 
 
 AprilS 
 
 To sundries, 
 
 1 
 
 60 
 
 10 
 
 1827. Dr. Richard Lewis, (York.) 
 
 April 8 
 24 
 
 To sundries, - 
 To sundries, ... 
 
 1 
 
 2 
 
 24 
 29 
 
 25 
 92 
 
 i 
 
 
 1827. Dr. George Simpson, 
 
 June 1 
 6 
 
 To balance from folio 1, 
 To wine, .... 
 
 2 
 
 1 9 
 12 
 
 I 
 
 00 
 00 
 
(2) 235 
 1827. Contra Cr. 
 
 April 6 
 
 By balance, 
 
 1 
 
 $ 
 90 
 
 as. 
 
 00 
 
 1827. Contra Cr. 
 
 
 
 
 
 
 1827. Contra Cr. 
 
 April 8 
 
 By cash, 
 
 1 
 
 15 
 
 00 
 
 1827. Contra Cr. 
 
 
 
 
 
 
236 TABLES. 
 
 A TABLE jor reducing Shillings and Pence into 
 and Mills. 
 
 SMI. Shil. Shil Shil. ShiL 
 
 
 
 
 
 1 
 
 
 
 2 
 
 3 
 
 4 
 
 
 5 
 
 Pn.cts.m. 
 
 cts. 
 
 m. 
 
 cts. 
 
 m. 
 
 cts. m. cts. 
 
 m. 
 
 cts. 
 
 m. 
 
 
 
 
 
 16 
 
 7 
 
 .33 
 
 3J50 
 
 0|66 
 
 7 
 
 83 
 
 3 
 
 1 
 
 1 
 
 4 
 
 18 
 
 1|34 
 
 7 
 
 51 
 
 468 
 
 1 
 
 84 
 
 7 
 
 2 
 
 2 
 
 8 
 
 19 
 
 536 
 
 1 
 
 52 
 
 8|69 
 
 5 
 
 86 
 
 1 
 
 3 
 
 4 
 
 2 
 
 20 
 
 9|37 
 
 5 
 
 54 
 
 2|70 
 
 9|87 
 
 5 
 
 4 
 
 5 
 
 6 
 
 22 
 
 338 
 
 9|55 
 
 6|72 
 
 388 
 
 9 
 
 5 
 
 7 
 
 
 
 23 
 
 7J40 
 
 3 
 
 57 
 
 0|73 
 
 7|90 
 
 3 
 
 6 
 
 8 
 
 3|25 
 
 0|41 
 
 6(58 
 
 3|75 
 
 0|91 
 
 6 
 
 7 
 
 9 
 
 7|26 
 
 4 
 
 43 
 
 059 
 
 7|76 
 
 4 
 
 93 
 
 
 
 8 
 
 11 
 
 1|27 
 
 8 
 
 44 
 
 4|61 
 
 1|77 
 
 8 
 
 94 
 
 4 
 
 9 
 
 12 
 
 o 
 
 29 
 
 
 
 45 
 
 8|62 
 
 5|79 
 
 295 
 
 8 
 
 10 
 
 13 
 
 9 
 
 30 
 
 6 
 
 47 
 
 2|63 
 
 980 
 
 6|97 
 
 2 
 
 11 
 
 15 
 
 332 
 
 0|48 
 
 6|65 
 
 3|82 
 
 0|98 
 
 6 
 
 EXAMPLE - Reduce 3s. 6d. to cents and mills. Look for 3s. 
 at the head of the column, and 6 under pence at the left hand 
 side : then casting- your eye along- in that line until you come te 
 the Ssi column, you have 58 cents 3 mills, the answer. 
 
 Tables of the value of the Gold Coins of Great Britain^Franct and 
 Spain according to the act of Congress of April 29, 1816. 
 
 [Gold Coins of FRANCE.] [Gold Coins of SPAINJ 
 
 grjct. grs.ct. pwt. % ct. pwt. $ cL | gr. ct gr. ct. pt. g, ct.pt. g ct. 
 
 f> 
 
 3 
 
 13 
 
 47 
 
 ] 
 
 0,87 
 
 
 11 
 
 9 
 
 ,60 I 
 
 1 
 
 3 
 
 13 
 
 45 
 
 1 0.84 
 
 11 
 
 9,24 
 
 2 
 
 7 
 
 14 
 
 51 
 
 2 
 
 1,75 
 
 
 12 
 
 10 
 
 > 47 1 
 
 2 
 
 7 
 
 14 
 
 49 
 
 2 
 
 1,68 
 
 12 
 
 10,08 
 
 3 
 
 11 
 
 15 
 
 55 
 
 3 
 
 262 
 
 
 13 
 
 11 
 
 ,34 
 
 3 
 
 11 
 
 15 
 
 52 
 
 3 
 
 2,52 
 
 13 
 
 10,92 
 
 4 
 
 14 
 
 16 
 
 58 
 
 4 
 
 3,49 
 
 
 14 
 
 12 
 
 ,21 f 
 
 4 
 
 14 
 
 16 
 
 56 
 
 4 
 
 3,36 
 
 14 
 
 11,76 
 
 5 
 
 18 
 
 17 
 
 62 
 
 5 
 
 4,36 
 
 
 15 
 
 13,09 { 
 
 5 
 
 17 
 
 17 
 
 59 
 
 5 
 
 4,20 
 
 15 
 
 12,60 
 
 6 
 
 22 
 
 18 
 
 65 
 
 6 
 
 3,23 
 
 
 16 
 
 13 
 
 : 96{ 
 
 6 
 
 >I 
 
 IS 
 
 f>3 
 
 6 
 
 5,04 
 
 16 
 
 13,44 
 
 7 
 
 25 
 
 19 
 
 68 
 
 7 
 
 6,11 
 
 
 17 
 
 14,83 < 
 
 7 
 
 24119 
 
 66 
 
 7 
 
 5,88 
 
 17 
 
 14,28 
 
 8 
 
 29 
 
 20 
 
 72 
 
 8 
 
 6,98 
 
 
 18 
 
 15,71 } 
 
 8 
 
 2820 
 
 70 
 
 8 
 
 6,72 
 
 18 
 
 15,12 
 
 9 
 
 33 
 
 21 
 
 76 
 
 9 
 
 7,85 
 
 
 19 
 
 16,58 \ 
 
 9 
 
 32 
 
 21 
 
 73 
 
 9 
 
 7,56 
 
 19 
 
 15,96 
 
 10 
 
 3,6 
 
 2280 
 
 10|8,73 
 
 
 20 
 
 17 
 
 ,45 * 
 
 10 
 
 36 
 
 22 
 
 77 
 
 108,402016,80 
 
 11 
 
 40 
 
 23 
 
 84 
 
 
 
 i 
 
 1 | 
 
 39 
 
 23 
 
 SO 
 
 
 12 
 
 44 
 
 
 
 
 1 
 
 12 
 
 42 
 
 
 
 
 Gold Coins of GREAT BRITAIN and PORTUGAL. 
 
 or 
 
 ct. 
 
 g-r 
 
 ct g 
 
 r ct. 
 
 gr 
 
 .ct. 
 
 pwt 
 
 dl.ct 
 
 pwt 
 
 dl. ct 
 
 spt 
 
 dl. cts 
 
 pwt 
 
 diets. 
 
 & 
 1 
 
 3 
 
 7 
 
 25 1 
 
 348 
 
 19170 
 
 I 
 
 0,89 
 
 1 
 
 6,2^ 
 
 \ \l 
 
 11,55 
 
 17 
 
 15,11 
 
 2 
 
 7 
 
 8 
 
 29 1 
 
 451 
 
 20174 
 
 2 
 
 1,78 
 
 6 
 
 7,11 
 
 \ 
 
 t 12,44 
 
 18 
 
 16,00 
 
 3 
 
 11 
 
 9 
 
 33 | 
 
 555 
 
 21 
 
 78 
 
 3 |2,67 fl 
 
 8,0( 
 
 )ii 
 
 > 13,33 19 
 
 16,89 
 
 4 
 
 If 
 
 10 
 
 371 
 
 659 
 
 22 
 
 81 
 
 4 
 
 3,55 
 
 1C 
 
 8,8< 
 
 ne 
 
 > 14,22 
 
 20 
 
 17,78. 
 
 
 18 
 
 11 
 
 40 1 
 
 763 
 
 23 
 
 85 
 
 5 
 
 4,44 
 
 11 
 
 9,78 
 
 
 6 
 
 22 12J44I18 67 
 
 
 *6 
 
 5,33 
 
 12 
 
 10,67 
 
 
FORMS OF NOTES AND BILLS OF EXCHANGE. 237 
 
 FORMS OF NOTES, BILLS, RECEIPTS, ifc. 
 
 PROMISSORY NOTE. 
 
 Hallowell, June 6, 1827. 
 
 FOR value received, I promise to pay one hundred and twen- 
 ty-one dollars and fifty cents to George Rich, or order, in sixty 
 days, with interest. HENRY WEST. 
 
 $121,50 
 
 Witness, Geo. Spelman. 
 
 I 
 
 PROMISSORY NOTE BY TWO PERSONS. 
 
 Hallowell, June 6, 1827. 
 
 For value received, we jointly and severally promise to pay fif- 
 ty-six dollars to A. B. or order, on demand, with interest. 
 
 C. Davis. 
 
 $56,00 E. Fox. 
 Attest, G, Hill, 
 
 NOTE FOR BORROWED MONEY. 
 Borrowed and received of C. D. forty-nine dollars, which I 
 promise to pay on demand. E. Fox. 
 
 $49,00 
 
 0^7" A promissory note having- order inserted, may be endorsed 
 from one person to another ; and if value received is not mention- 
 ed, it is of no force. 
 
 INLAND BILL OF EXCHANGE. 
 
 $1000,00 Portland, June 6, 1 827. 
 
 Ten days after sight, pay to George Brown or order, one thou- 
 sand dollars, for value received, and place it to my account with- 
 out further advice, (or as advised,) from 
 
 Your humble servant, HENRT WEST. 
 
 To Mr. George Rich, Boston. 
 
 FOREIGN BILL OF EXCHANGE. 
 
 EXCHANGE for 400 sterling. 
 
 Hallowell, June 7, 1827. 
 
 Sixty days after sight, (or at usance,*) pay this my first bill of 
 exchange, second and third of the same tenor and date not paid, 
 to Mr. George Brown or his order, four hundred pounds sterling 
 
 * Usance is a customary time for the payment of foreign bills of 
 exchange,circulating from one nation to another; and varies from 
 80 to 90 days, according to the custom of different countries. 
 
238 RECEIPTS BANK DISCOUNT. 
 
 (exchange at four shilling's and sixpence per dollar) for value re- 
 ceived, and place it (with or without further advice,) to the ac- 
 count of Your humble servant. 
 
 HENRY WEST. 
 Messrs. Neil fy Thompson, 
 Merchants, Liverpool. 
 
 RECEIPT FOR MONEY PAID ON NOTE, 
 
 Hallowell.Dec 6, 1827. Received from William Grant (by the 
 hands of Thomas Amory) sixty-one dollars and fifty cents, which 
 is endorsed on his note of May 16th. 1825. 
 
 SAMUEL PRINCE. 
 $61,50 
 
 RECEIPT FOR MONEY RECEIVED ON ACCOUNT. 
 
 June 7, 1827 Received from D. E. (by the hands of G H.) 
 forty dollars on account. L. M. 
 
 $40,00 
 
 GENERAL RECEIPT. 
 
 June 7, 1827. Received of N. O. ten dollars and twenty-nine 
 cents, in full of all demands. N. B. 
 
 $10,29 
 
 N. B A general receipt will discharge all debts, except such 
 as are on specialty, that is, bonds, bills, and other instruments 
 that may properly be called acts or deeds, viz, those that require 
 to be executed in a solemn manner, where the sealing and deliv- 
 ery are the most essential parts of the act, and on that account 
 can only be destroyed or cancelled by something of equal force, 
 viz, some other specialty, such as a general release, &c. Neither 
 wUl it discharge endorseable promissory notes, or Inland bills. 
 
 BANK DISCOUNT. 
 
 When a note is offered at a bank for discount, two endorsers are 
 generally required, to the first of whom it is made payable : Thus 
 A, having occasion to borrow money, procures B and C as en- 
 dorsers to nig note, and offers it for discount in the following form. 
 
 $500,00 Hallowell, June 6, 1827. 
 
 For value received, I promise fro pay five hundred dollars toB 
 or order, at the^Gardiner Bank, in fifty -seven days, with custom- 
 ary grace. 
 
 The method used among bankers in discounting notes, &c is to 
 
INVOICE AND ACCOUNT. 239 
 
 find the interest of the sum from the date of the note to the time 
 when it becomes due, including the days of grace ; the interest 
 thus found is reckoned the discount, and is taken from the amount 
 of the note at a time, before the person receives his money. 
 
 Grace denotes a term of three days, which custom has allowed 
 to the borrower : that is, though the note becomes due in fifty- 
 seven days, he may withhold payment until the sixtieth, for which 
 reason the interest is reckoned for sixty days, notwithstanding 
 tne note should be paid the fifty-seventh day. 
 
 INVOICE OF GOODS. 
 
 Boston, June 6, 1827. 
 Mr. N. BROWN bought of 
 
 GEORGE RICH. 
 
 32 ells Mode, at 3s. 4d. $17 78 
 
 64 yds. Striped Nankins, - - Is. 6d. 16 00 
 
 28 *' Calico, Is. 9d. 6 17 
 
 4 pieces Muslin, < 30s. - 20 00 
 
 56 yds. Cotton Cassimere, - 2s. - - 18 67 
 
 20 pieces of India Cottons, - - 18s. - 60 00 
 
 25 " plain Nankins, - 6s. 6d. - 27 08 
 
 2 doz. cotton Hose, (Metis) 66s. - 22 00 
 
 $189 70 
 Rec'd payment by his Note at 60 days, 
 
 GEO. HIGH. 
 
 ACCOUNT RENDERED. 
 
 MR. RICHARD LEWIS, 
 1827. To A- 
 
 April 1, To 2yds. Superfine Cloth, 
 
 4 " blk. Cassimere, 
 
 1J doz. Buttons, 
 
 1 small do. 
 
 4 skeins silk, 
 
 4 sticks of twist, 
 24, 5 pieces Ind. Cottons, 22yds. ea. 
 
 7 yds. Cotton Cambric, 
 
 6 " col'd do. 
 May 18, 14 do. do. 
 
 7 " Linen, 
 
 1 oz. Thread, No. 40, - * 
 
 2 pair Morocco Shoes, - 1,08 
 1 oz. Indigo, 
 
 Hallowell, Dec. 6, 1827. $70 43 
 
 Rec'd payment for A B . 
 
 GEORGE NORTH. 
 
240 
 
 ACCOUNT CURRENT. 
 
 
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